If $a_n > 0$ prove that $sum_{n=1}^{infty} frac{a_n}{(a_1+1)(a_2+1)cdots(a_n+1)}$ converges [duplicate]
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This question already has an answer here:
Prove the convergence of a series.
1 answer
I have an interesting task: If $a_n > 0$, prove that $$sum_{n=1}^{infty} frac{a_n}{(a_1+1)(a_2+1)cdots(a_n+1)}$$ converges.
I thought that it will be simple because ratio test gives me:
$$frac{u_{n+1}}{u_n}= frac{a_{n+1}}{a_{n+1}+1}cdot a_n^{-1} < 1 cdot a_n^{-1} = frac{1}{a_n}$$ and $a_n$ should be in $[0,1]$. But... In my opinion it can be over that... why need I assume that $ a_n rightarrow g in [0,1] $?
There is similar topic on this forum, but It was not solved there...
@edit
I saw that:
$$sum_{n=1}^{N}frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-frac{1}{(1+a_1)(1+a_2)...(1+a_N)} < 1 $$
So if series of partial sum is bounded from up, the sum converges, that is right?
@edit2 but It is good? Look at that:
$$ sum_{n=1}^{N}frac{a_n+1-1}{(1+a_1)(1+a_2)...(1+a_n)} = sum_{n=1}^{N}frac{1}{(1+a_1)(1+a_2)...(1+a_{n-1})}-frac{1}{(1+a_1)(1+a_2)...(1+a_n)} $$ why somebody changed first part into $1$?
@edit3 Ok, I think that I have understood, thanks for your time ;)
real-analysis convergence
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
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Dec 17 '18 at 18:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 1 more comment
$begingroup$
This question already has an answer here:
Prove the convergence of a series.
1 answer
I have an interesting task: If $a_n > 0$, prove that $$sum_{n=1}^{infty} frac{a_n}{(a_1+1)(a_2+1)cdots(a_n+1)}$$ converges.
I thought that it will be simple because ratio test gives me:
$$frac{u_{n+1}}{u_n}= frac{a_{n+1}}{a_{n+1}+1}cdot a_n^{-1} < 1 cdot a_n^{-1} = frac{1}{a_n}$$ and $a_n$ should be in $[0,1]$. But... In my opinion it can be over that... why need I assume that $ a_n rightarrow g in [0,1] $?
There is similar topic on this forum, but It was not solved there...
@edit
I saw that:
$$sum_{n=1}^{N}frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-frac{1}{(1+a_1)(1+a_2)...(1+a_N)} < 1 $$
So if series of partial sum is bounded from up, the sum converges, that is right?
@edit2 but It is good? Look at that:
$$ sum_{n=1}^{N}frac{a_n+1-1}{(1+a_1)(1+a_2)...(1+a_n)} = sum_{n=1}^{N}frac{1}{(1+a_1)(1+a_2)...(1+a_{n-1})}-frac{1}{(1+a_1)(1+a_2)...(1+a_n)} $$ why somebody changed first part into $1$?
@edit3 Ok, I think that I have understood, thanks for your time ;)
real-analysis convergence
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
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marked as duplicate by JimmyK4542, Winther, jgon, Did real-analysis
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Dec 17 '18 at 18:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You shouldn't be using the ratio test at all.
$endgroup$
– user10354138
Dec 17 '18 at 18:28
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I suppose you must assume $a_0=0$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:33
$begingroup$
It is not finished there
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– VirtualUser
Dec 17 '18 at 18:34
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@user10354138 I know, but there it should works too
$endgroup$
– VirtualUser
Dec 17 '18 at 18:35
1
$begingroup$
@VirtualUser: Since sequence of partial sums are monotonic and bounded (from above), so the partial sums converge to the supremum as a consequence of Monotone convergence theorem. Hence the summation converges to the supremum.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:42
|
show 1 more comment
$begingroup$
This question already has an answer here:
Prove the convergence of a series.
1 answer
I have an interesting task: If $a_n > 0$, prove that $$sum_{n=1}^{infty} frac{a_n}{(a_1+1)(a_2+1)cdots(a_n+1)}$$ converges.
I thought that it will be simple because ratio test gives me:
$$frac{u_{n+1}}{u_n}= frac{a_{n+1}}{a_{n+1}+1}cdot a_n^{-1} < 1 cdot a_n^{-1} = frac{1}{a_n}$$ and $a_n$ should be in $[0,1]$. But... In my opinion it can be over that... why need I assume that $ a_n rightarrow g in [0,1] $?
There is similar topic on this forum, but It was not solved there...
@edit
I saw that:
$$sum_{n=1}^{N}frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-frac{1}{(1+a_1)(1+a_2)...(1+a_N)} < 1 $$
So if series of partial sum is bounded from up, the sum converges, that is right?
@edit2 but It is good? Look at that:
$$ sum_{n=1}^{N}frac{a_n+1-1}{(1+a_1)(1+a_2)...(1+a_n)} = sum_{n=1}^{N}frac{1}{(1+a_1)(1+a_2)...(1+a_{n-1})}-frac{1}{(1+a_1)(1+a_2)...(1+a_n)} $$ why somebody changed first part into $1$?
@edit3 Ok, I think that I have understood, thanks for your time ;)
real-analysis convergence
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
$endgroup$
This question already has an answer here:
Prove the convergence of a series.
1 answer
I have an interesting task: If $a_n > 0$, prove that $$sum_{n=1}^{infty} frac{a_n}{(a_1+1)(a_2+1)cdots(a_n+1)}$$ converges.
I thought that it will be simple because ratio test gives me:
$$frac{u_{n+1}}{u_n}= frac{a_{n+1}}{a_{n+1}+1}cdot a_n^{-1} < 1 cdot a_n^{-1} = frac{1}{a_n}$$ and $a_n$ should be in $[0,1]$. But... In my opinion it can be over that... why need I assume that $ a_n rightarrow g in [0,1] $?
There is similar topic on this forum, but It was not solved there...
@edit
I saw that:
$$sum_{n=1}^{N}frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-frac{1}{(1+a_1)(1+a_2)...(1+a_N)} < 1 $$
So if series of partial sum is bounded from up, the sum converges, that is right?
@edit2 but It is good? Look at that:
$$ sum_{n=1}^{N}frac{a_n+1-1}{(1+a_1)(1+a_2)...(1+a_n)} = sum_{n=1}^{N}frac{1}{(1+a_1)(1+a_2)...(1+a_{n-1})}-frac{1}{(1+a_1)(1+a_2)...(1+a_n)} $$ why somebody changed first part into $1$?
@edit3 Ok, I think that I have understood, thanks for your time ;)
This question already has an answer here:
Prove the convergence of a series.
1 answer
real-analysis convergence
real-analysis convergence
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
edited Dec 17 '18 at 18:46
VirtualUser
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
asked Dec 17 '18 at 18:15
VirtualUserVirtualUser
1,096117
1,096117
marked as duplicate by JimmyK4542, Winther, jgon, Did real-analysis
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Dec 17 '18 at 18:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by JimmyK4542, Winther, jgon, Did real-analysis
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Dec 17 '18 at 18:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
You shouldn't be using the ratio test at all.
$endgroup$
– user10354138
Dec 17 '18 at 18:28
$begingroup$
I suppose you must assume $a_0=0$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:33
$begingroup$
It is not finished there
$endgroup$
– VirtualUser
Dec 17 '18 at 18:34
$begingroup$
@user10354138 I know, but there it should works too
$endgroup$
– VirtualUser
Dec 17 '18 at 18:35
1
$begingroup$
@VirtualUser: Since sequence of partial sums are monotonic and bounded (from above), so the partial sums converge to the supremum as a consequence of Monotone convergence theorem. Hence the summation converges to the supremum.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:42
|
show 1 more comment
1
$begingroup$
You shouldn't be using the ratio test at all.
$endgroup$
– user10354138
Dec 17 '18 at 18:28
$begingroup$
I suppose you must assume $a_0=0$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:33
$begingroup$
It is not finished there
$endgroup$
– VirtualUser
Dec 17 '18 at 18:34
$begingroup$
@user10354138 I know, but there it should works too
$endgroup$
– VirtualUser
Dec 17 '18 at 18:35
1
$begingroup$
@VirtualUser: Since sequence of partial sums are monotonic and bounded (from above), so the partial sums converge to the supremum as a consequence of Monotone convergence theorem. Hence the summation converges to the supremum.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:42
1
1
$begingroup$
You shouldn't be using the ratio test at all.
$endgroup$
– user10354138
Dec 17 '18 at 18:28
$begingroup$
You shouldn't be using the ratio test at all.
$endgroup$
– user10354138
Dec 17 '18 at 18:28
$begingroup$
I suppose you must assume $a_0=0$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:33
$begingroup$
I suppose you must assume $a_0=0$.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:33
$begingroup$
It is not finished there
$endgroup$
– VirtualUser
Dec 17 '18 at 18:34
$begingroup$
It is not finished there
$endgroup$
– VirtualUser
Dec 17 '18 at 18:34
$begingroup$
@user10354138 I know, but there it should works too
$endgroup$
– VirtualUser
Dec 17 '18 at 18:35
$begingroup$
@user10354138 I know, but there it should works too
$endgroup$
– VirtualUser
Dec 17 '18 at 18:35
1
1
$begingroup$
@VirtualUser: Since sequence of partial sums are monotonic and bounded (from above), so the partial sums converge to the supremum as a consequence of Monotone convergence theorem. Hence the summation converges to the supremum.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:42
$begingroup$
@VirtualUser: Since sequence of partial sums are monotonic and bounded (from above), so the partial sums converge to the supremum as a consequence of Monotone convergence theorem. Hence the summation converges to the supremum.
$endgroup$
– Yadati Kiran
Dec 17 '18 at 18:42
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
$$
eqalign{
& sumlimits_{1, le ,n,} {{{a_n } over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}} = cr
& = sumlimits_{1, le ,n,} {{{left( {a_n + 1} right) - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}}
= cdots cr}
$$
(continuing)
$$
eqalign{
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}}
- {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = 1 - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cdots cr}
$$
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
I used your hint to solve my problem, can you check if I done this well?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:39
$begingroup$
yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude.
$endgroup$
– G Cab
Dec 17 '18 at 22:53
add a comment |
$begingroup$
Here, the ratio test is useless because you have zero information on $a_n$.
May I suggest that you compute the first partial sums to “get a feeling” about what happens?
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$$
eqalign{
& sumlimits_{1, le ,n,} {{{a_n } over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}} = cr
& = sumlimits_{1, le ,n,} {{{left( {a_n + 1} right) - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}}
= cdots cr}
$$
(continuing)
$$
eqalign{
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}}
- {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = 1 - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cdots cr}
$$
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
I used your hint to solve my problem, can you check if I done this well?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:39
$begingroup$
yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude.
$endgroup$
– G Cab
Dec 17 '18 at 22:53
add a comment |
$begingroup$
Hint:
$$
eqalign{
& sumlimits_{1, le ,n,} {{{a_n } over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}} = cr
& = sumlimits_{1, le ,n,} {{{left( {a_n + 1} right) - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}}
= cdots cr}
$$
(continuing)
$$
eqalign{
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}}
- {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = 1 - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cdots cr}
$$
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
I used your hint to solve my problem, can you check if I done this well?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:39
$begingroup$
yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude.
$endgroup$
– G Cab
Dec 17 '18 at 22:53
add a comment |
$begingroup$
Hint:
$$
eqalign{
& sumlimits_{1, le ,n,} {{{a_n } over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}} = cr
& = sumlimits_{1, le ,n,} {{{left( {a_n + 1} right) - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}}
= cdots cr}
$$
(continuing)
$$
eqalign{
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}}
- {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = 1 - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cdots cr}
$$
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
$endgroup$
Hint:
$$
eqalign{
& sumlimits_{1, le ,n,} {{{a_n } over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}} = cr
& = sumlimits_{1, le ,n,} {{{left( {a_n + 1} right) - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right) cdots left( {a_n + 1} right)}}}
= cdots cr}
$$
(continuing)
$$
eqalign{
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 } over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {{a_2 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {{a_3 + 1 - 1} over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}}
+ {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)}} - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = {{a_1 } over {left( {a_1 + 1} right)}} + {1 over {left( {a_1 + 1} right)}}
- {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cr
& = 1 - {1 over {left( {a_1 + 1} right)left( {a_2 + 1} right)left( {a_3 + 1} right)}} + cdots = cdots cr}
$$
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
edited Dec 17 '18 at 22:49
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
answered Dec 17 '18 at 18:30
G CabG Cab
20.4k31341
20.4k31341
$begingroup$
I used your hint to solve my problem, can you check if I done this well?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:39
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yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude.
$endgroup$
– G Cab
Dec 17 '18 at 22:53
add a comment |
$begingroup$
I used your hint to solve my problem, can you check if I done this well?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:39
$begingroup$
yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude.
$endgroup$
– G Cab
Dec 17 '18 at 22:53
$begingroup$
I used your hint to solve my problem, can you check if I done this well?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:39
$begingroup$
I used your hint to solve my problem, can you check if I done this well?
$endgroup$
– VirtualUser
Dec 17 '18 at 18:39
$begingroup$
yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude.
$endgroup$
– G Cab
Dec 17 '18 at 22:53
$begingroup$
yes, you got the idea (it's a telescoping sum), but you shall pay attention to the starting point (there is not an $a_0$ to subtract): I continued for some further steps .. now you shall be able to conclude.
$endgroup$
– G Cab
Dec 17 '18 at 22:53
add a comment |
$begingroup$
Here, the ratio test is useless because you have zero information on $a_n$.
May I suggest that you compute the first partial sums to “get a feeling” about what happens?
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
$endgroup$
add a comment |
$begingroup$
Here, the ratio test is useless because you have zero information on $a_n$.
May I suggest that you compute the first partial sums to “get a feeling” about what happens?
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
$endgroup$
add a comment |
$begingroup$
Here, the ratio test is useless because you have zero information on $a_n$.
May I suggest that you compute the first partial sums to “get a feeling” about what happens?
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
$endgroup$
Here, the ratio test is useless because you have zero information on $a_n$.
May I suggest that you compute the first partial sums to “get a feeling” about what happens?
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
answered Dec 17 '18 at 18:30
MindlackMindlack
4,920211
4,920211
add a comment |
add a comment |
1
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You shouldn't be using the ratio test at all.
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– user10354138
Dec 17 '18 at 18:28
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I suppose you must assume $a_0=0$.
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– Yadati Kiran
Dec 17 '18 at 18:33
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It is not finished there
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– VirtualUser
Dec 17 '18 at 18:34
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@user10354138 I know, but there it should works too
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– VirtualUser
Dec 17 '18 at 18:35
1
$begingroup$
@VirtualUser: Since sequence of partial sums are monotonic and bounded (from above), so the partial sums converge to the supremum as a consequence of Monotone convergence theorem. Hence the summation converges to the supremum.
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– Yadati Kiran
Dec 17 '18 at 18:42