$sum_{i=1}^{infty}frac{1}{a_i}$ converges.,prove that $lim_{n to infty}frac{b_n}{n}=0$.
Help: Let $a_1,a_2,...$ be positive integers such that $sum_{i=1}^{infty}frac{1}{a_i}$ converges. For each $n$, let $b_n$ denote the number of positive integers $i$ for which $a_i leq n$. prove that $lim_{n to infty}frac{b_n}{n}=0$.
Intuitively, $a_i$ should grow large fast enough for $1/a_i$ to converge. so gap between $a_i$ should be bigger and bigger? Is this the right idea?
calculus summation contest-math
asked Nov 23 at 6:22
mathnoob
1,761422
add a comment |
Help: Let $a_1,a_2,...$ be positive integers such that $sum_{i=1}^{infty}frac{1}{a_i}$ converges. For each $n$, let $b_n$ denote the number of positive integers $i$ for which $a_i leq n$. prove that $lim_{n to infty}frac{b_n}{n}=0$.
Intuitively, $a_i$ should grow large fast enough for $1/a_i$ to converge. so gap between $a_i$ should be bigger and bigger? Is this the right idea?
calculus summation contest-math
asked Nov 23 at 6:22
mathnoob
1,761422
1
You have tagged this "contest-math". Which contest, please?
– Gerry Myerson
Nov 23 at 6:30
1
Putnam contest 1964 B1
– mathnoob
Nov 23 at 6:31
I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky.
– Jimmy Sabater
Nov 23 at 6:32
add a comment |
Help: Let $a_1,a_2,...$ be positive integers such that $sum_{i=1}^{infty}frac{1}{a_i}$ converges. For each $n$, let $b_n$ denote the number of positive integers $i$ for which $a_i leq n$. prove that $lim_{n to infty}frac{b_n}{n}=0$.
Intuitively, $a_i$ should grow large fast enough for $1/a_i$ to converge. so gap between $a_i$ should be bigger and bigger? Is this the right idea?
calculus summation contest-math
asked Nov 23 at 6:22
mathnoob
1,761422
Help: Let $a_1,a_2,...$ be positive integers such that $sum_{i=1}^{infty}frac{1}{a_i}$ converges. For each $n$, let $b_n$ denote the number of positive integers $i$ for which $a_i leq n$. prove that $lim_{n to infty}frac{b_n}{n}=0$.
Intuitively, $a_i$ should grow large fast enough for $1/a_i$ to converge. so gap between $a_i$ should be bigger and bigger? Is this the right idea?
calculus summation contest-math
calculus summation contest-math
asked Nov 23 at 6:22
mathnoob
1,761422
asked Nov 23 at 6:22
mathnoob
1,761422
edited Nov 23 at 6:33
asked Nov 23 at 6:22
mathnoob
1,761422
asked Nov 23 at 6:22
mathnoob
1,761422
asked Nov 23 at 6:22
mathnoob
1,761422
1,761422
1
You have tagged this "contest-math". Which contest, please?
– Gerry Myerson
Nov 23 at 6:30
1
Putnam contest 1964 B1
– mathnoob
Nov 23 at 6:31
I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky.
– Jimmy Sabater
Nov 23 at 6:32
add a comment |
1
You have tagged this "contest-math". Which contest, please?
– Gerry Myerson
Nov 23 at 6:30
1
Putnam contest 1964 B1
– mathnoob
Nov 23 at 6:31
I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky.
– Jimmy Sabater
Nov 23 at 6:32
1
1
You have tagged this "contest-math". Which contest, please?
– Gerry Myerson
Nov 23 at 6:30
You have tagged this "contest-math". Which contest, please?
– Gerry Myerson
Nov 23 at 6:30
1
1
Putnam contest 1964 B1
– mathnoob
Nov 23 at 6:31
Putnam contest 1964 B1
– mathnoob
Nov 23 at 6:31
I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky.
– Jimmy Sabater
Nov 23 at 6:32
I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky.
– Jimmy Sabater
Nov 23 at 6:32
add a comment |
1 Answer
1
active
oldest
votes
Let $epsilon >0$ and choose $N$ such that $sum_{i=N}^{infty} frac 1 {a_i} <epsilon$. Let $J_n={i>N:a_i leq n}$. Then $sum_{iin J_n} frac 1 nleq sum_{iin J_n} frac 1 {a_i} <epsilon$. Hence $frac {card(J_n)} n <epsilon$ for all $n$. Now $b_n leq N+card(J_n)$ so $frac {b_n} n leq epsilon +frac N n to epsilon $ as $n to infty$.
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
add a comment |
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Let $epsilon >0$ and choose $N$ such that $sum_{i=N}^{infty} frac 1 {a_i} <epsilon$. Let $J_n={i>N:a_i leq n}$. Then $sum_{iin J_n} frac 1 nleq sum_{iin J_n} frac 1 {a_i} <epsilon$. Hence $frac {card(J_n)} n <epsilon$ for all $n$. Now $b_n leq N+card(J_n)$ so $frac {b_n} n leq epsilon +frac N n to epsilon $ as $n to infty$.
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
add a comment |
Let $epsilon >0$ and choose $N$ such that $sum_{i=N}^{infty} frac 1 {a_i} <epsilon$. Let $J_n={i>N:a_i leq n}$. Then $sum_{iin J_n} frac 1 nleq sum_{iin J_n} frac 1 {a_i} <epsilon$. Hence $frac {card(J_n)} n <epsilon$ for all $n$. Now $b_n leq N+card(J_n)$ so $frac {b_n} n leq epsilon +frac N n to epsilon $ as $n to infty$.
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
add a comment |
Let $epsilon >0$ and choose $N$ such that $sum_{i=N}^{infty} frac 1 {a_i} <epsilon$. Let $J_n={i>N:a_i leq n}$. Then $sum_{iin J_n} frac 1 nleq sum_{iin J_n} frac 1 {a_i} <epsilon$. Hence $frac {card(J_n)} n <epsilon$ for all $n$. Now $b_n leq N+card(J_n)$ so $frac {b_n} n leq epsilon +frac N n to epsilon $ as $n to infty$.
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
Let $epsilon >0$ and choose $N$ such that $sum_{i=N}^{infty} frac 1 {a_i} <epsilon$. Let $J_n={i>N:a_i leq n}$. Then $sum_{iin J_n} frac 1 nleq sum_{iin J_n} frac 1 {a_i} <epsilon$. Hence $frac {card(J_n)} n <epsilon$ for all $n$. Now $b_n leq N+card(J_n)$ so $frac {b_n} n leq epsilon +frac N n to epsilon $ as $n to infty$.
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
answered Nov 23 at 6:36
Kavi Rama Murthy
48.9k31854
48.9k31854
add a comment |
add a comment |
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1
You have tagged this "contest-math". Which contest, please?
– Gerry Myerson
Nov 23 at 6:30
1
Putnam contest 1964 B1
– mathnoob
Nov 23 at 6:31
I was thinking on using the squeeze, but upon reading the question more carefully, it is actually tricky.
– Jimmy Sabater
Nov 23 at 6:32