$sum_{i=1}^{infty}frac{1}{a_i}$ converges.,prove that $lim_{n to infty}frac{b_n}{n}=0$.












3














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







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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
















3














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







share|cite|improve this question




















  • 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














3












3








3


1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 6:33

























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














  • 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










1 Answer
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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 $n to infty$.






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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 $n to infty$.






    share|cite|improve this answer


























      4














      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 $n to infty$.






      share|cite|improve this answer
























        4












        4








        4






        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 $n to infty$.






        share|cite|improve this answer












        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 $n to infty$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 at 6:36









        Kavi Rama Murthy

        48.9k31854




        48.9k31854






























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