0
$\begingroup$

Possible Duplicate:
Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?

I need to Prove the following sum converges:

$$\lim_{n\to\infty}\sum\limits_{i=1}^n\dfrac{1}{n+i}$$

What methods can I use?

  • 0
    Comparison of sums and integrals.2012-09-29
  • 0
    Are you taking the limit for $n\to \infty$?2012-09-29
  • 0
    Yes, I will specify that2012-09-29
  • 2
    Language note: This is not a "sum convergence" problem. We say a sum converges of the sequence $a_0$, $a_0+a_1$, $a_0+a_1+a_2$,... converges. This problem is not a problem of that type. Rather this is a question about the limit of a sequence of different (related) sums.2012-09-29

4 Answers 4

0

$$\sum_{i=1}^n\frac{1}{n+i}=\frac{1}{n}\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\xrightarrow [n\to\infty]{} \int_0^1\frac{dx}{1+x}=...$$

  • 0
    Could you show some derivation of how you get $\int\dfrac{dx}{1+x}$2012-09-29
  • 0
    Riemann sums, as simple as that. Since the function $\,f(x)=\frac{1}{1+x}\,$ is continuous in $\,[0,1]\,$ its Riemann integral exists there and we can thus choose *any* subdivision of $\,[0,1]\,$ we want to form the Riemann sums. We choose $\,\{x_0=0\,,\,x_1=1/n\,,\,x_2=2/n\,,...,\,x_n=n/n=1\}\,$ and we form the Riemann sum $$\sum_{i=1}^nf(x_i)(x_i-x_{i-1})=\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\cdot\frac{1}{n}$$2012-09-29
4

Hint: you can rewrite your sum as $$ \lim_{n\to\infty} \frac{1-0}{n}\sum_{i=1}^n \frac{1}{1+\frac{i}{n}}, $$ Now do you know the definition of Riemann's integral?

Added: Somehow the english Wiki page doesn't not seem to show this as explicitly as the french one does, but you can have a look at this page, the first section shows what you have, with $f$ replaced by $\frac{1}{1+x}$

  • 0
    Is Riemann's integral that theorem they teach you in calc 1 when you take the number of rectangles under a curve to be infinite and compute their areas? If so, it has been a long long time! hahah2012-09-29
  • 0
    It is somewhat the way Riemann's integral is constructed in the first calculus course, via approximatio of areas under curves by rectangle.2012-09-29
3

HINT: The sequence of sums is bounded above by $1$: $$\sum_{i=1}^n\dfrac{1}{n+i}<\sum_{i=1}^n\frac1n=1\;.$$

It’s also strictly increasing, as you can show by calculating

$$\sum_{i=1}^{n+1}\dfrac{1}{n+1+i}-\sum_{i=1}^n\dfrac{1}{n+i}=\frac1{2(n+1)}+\sum_{i=1}^n\left(\frac1{n+1+i}-\frac1{n+i}\right)\;;$$

I’ll leave the rest of that calculation to you. Note that the last sum telescopes.

  • 0
    What does this last equation show? You are finding the $(n+1)th$ term correct?2012-09-29
  • 0
    @CodeKingPlusPlus: No, if $s_n$ is the $n$-th term, I’m calculating $s_{n+1}-s_n$. What kind of result would show that the sequence of $s_n$’s is increasing?2012-09-29
  • 0
    Ok, so if the difference of $s_{n+1} - s_n > 0$ then $s_n$ is increasing.2012-09-29
  • 0
    @CodeKingPlusPlus: Exactly.2012-09-29
0

Hint

$\frac{1}{n+i}=\frac{1}{n\left(1+\frac{i}{n}\right)}$, then rewrite sum $\sum\limits_{i=1}^n\dfrac{1}{n+i}$ as Riemann sum for appropriate integral.