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?
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?
$\sum_{i=1}^n\frac{1}{n+i}=\frac{1}{n}\sum_{i=1}^n\frac{1}{1+\frac{i}{n}}\xrightarrow [n\to\infty]{} \int_0^1\frac{dx}{1+x}=...$
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 $\frac{1}{1+x}$
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.
$\frac{1}{n+i}=\frac{1}{n\left(1+\frac{i}{n}\right)}$, then rewrite sum $\sum\limits_{i=1}^n\dfrac{1}{n+i}$ as Riemann sum for appropriate integral.