Convergence and divergence of series when changing finite number of summands
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I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:
Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have
begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and
begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}
for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.
Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.
real-analysis sequences-and-series convergence
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I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:
Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have
begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and
begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}
for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.
Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.
real-analysis sequences-and-series convergence
1
Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36
add a comment |
up vote
0
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up vote
0
down vote
favorite
I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:
Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have
begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and
begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}
for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.
Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.
real-analysis sequences-and-series convergence
I found a text saying that adding or changing only a finite number of summands does not have an effect on the convergence/divergence of the series. This is shown by the following argumentation:
Let $sum_{k=n_1}^{infty} a_k$ and $sum_{k=n_2}^{infty}b_k$ be two series with $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ their partial sums. Let's suppose there exists an $N$ so that $a_k = b_k$ for alle $kgeq N$, than we have
begin{align}s_n = sum_{k=n_1}^{n}a_k = a_{n_1} + a_{n_1+1} + ldots + a_{N-1} + sum_{k=N}^{n}a_kend{align} and
begin{align}t_n &= sum_{k=n_2}^{n}b_k = b_{n_2} + b_{n_2+1} + ldots + b_{N-1} + sum_{k=N}^{n}a_k \
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)end{align}
for all $n geq N$. Hence, both $(s_n)_{n geq n_1}$ and $(t_n)_ {tgeq n_2}$ are either convergent or divergent.
Unfortunately I do not see why $(s_n)_{n geq n_1}$ and $(t_n)_{t geq n_2}$ are either convergent or divergent following this calculation. Moreover I also don't get why this is showing that a finite number of changes to the summands of the series does not change the convergence behaviour of the series.
Can someone please help me understanding this proof.
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Nov 14 at 16:24
asked Nov 14 at 16:13
offline
547
547
1
Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36
add a comment |
1
Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36
1
1
Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36
Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36
add a comment |
1 Answer
1
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1
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accepted
Suppose $lim_{n to infty}t_nto t$,
and
begin{align}t_n
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}
We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have
$$t_n = s_n - C$$
then we have
begin{align}lim_{n to infty}t_n
&= lim_{n to infty}s_n - Cend{align}
Hence begin{align}
lim_{n to infty}s_n =t+ Cend{align}
That is $s_n$ converges as well.
Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose $lim_{n to infty}t_nto t$,
and
begin{align}t_n
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}
We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have
$$t_n = s_n - C$$
then we have
begin{align}lim_{n to infty}t_n
&= lim_{n to infty}s_n - Cend{align}
Hence begin{align}
lim_{n to infty}s_n =t+ Cend{align}
That is $s_n$ converges as well.
Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.
add a comment |
up vote
1
down vote
accepted
Suppose $lim_{n to infty}t_nto t$,
and
begin{align}t_n
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}
We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have
$$t_n = s_n - C$$
then we have
begin{align}lim_{n to infty}t_n
&= lim_{n to infty}s_n - Cend{align}
Hence begin{align}
lim_{n to infty}s_n =t+ Cend{align}
That is $s_n$ converges as well.
Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose $lim_{n to infty}t_nto t$,
and
begin{align}t_n
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}
We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have
$$t_n = s_n - C$$
then we have
begin{align}lim_{n to infty}t_n
&= lim_{n to infty}s_n - Cend{align}
Hence begin{align}
lim_{n to infty}s_n =t+ Cend{align}
That is $s_n$ converges as well.
Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.
Suppose $lim_{n to infty}t_nto t$,
and
begin{align}t_n
&= s_n - left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) + left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right).end{align}
We can let $left(a_{n_1} + a_{n_1+1} + ldots + a_{N-1}right) - left(b_{n_2} + b_{n_2+1} + ldots + b_{N-1}right)=C,$ a constant that is independent of $n$, then we have
$$t_n = s_n - C$$
then we have
begin{align}lim_{n to infty}t_n
&= lim_{n to infty}s_n - Cend{align}
Hence begin{align}
lim_{n to infty}s_n =t+ Cend{align}
That is $s_n$ converges as well.
Similarly, we can argue that if $s_n$ converges, then $t_n$ converges.
answered Nov 14 at 16:44
Siong Thye Goh
93.6k1462114
93.6k1462114
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Both partial series to $N-1$ contain a finite number of finite-valued terms, so these partial sums are finite. This finiteness will not affect whether the total sums (with the infinite number of terms) is finite or not.
– Lucozade
Nov 14 at 16:36