# Limit of a partial sum [closed]

I want to find the limit $$\lim_{n\to \infty}\sum_{i=1}^n \frac{1}{n+i}$$

I tried this. But I am not able to do it. Can anyone please help how to proceed?

• Its a Riemann-sum :) – ParaH2 Aug 28 '15 at 19:56
• Clicked on the wrong button and gone is the original comment...you might want to check out this question. – Hirshy Aug 28 '15 at 20:15

$$S_n=\sum_{i=1}^{n} \frac{1}{n+i}=\frac{1}{n}\sum_{i=1}^n \frac{1}{1+\frac{i}{n}}$$

Now let $f$ as : $f : x \mapsto\frac{1}{1+x}$

We can see that $$S_n=\frac{1}{n}\sum_{i=1}^{n} f\left(1+\frac{i}{n}\right)$$

Then $$\lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{i}{n}}=\int_0^1 \frac{\mathrm{d}x}{1+x}=\ln(2)$$

• Are you missing a factor of $\frac{1}{n}$ on the last line? After the limit and before the sum? – Fly by Night Aug 28 '15 at 21:09
• I corrected this :) – ParaH2 Aug 28 '15 at 21:12

Notice that

$\displaystyle\sum_{i=1}^n\dfrac{1}{n+i}=\sum_{i=1}^{2n}\dfrac1n-\sum_{i=1}^n\dfrac1n$

$\displaystyle\sum_{i=1}^n\dfrac1n=\ln n+\gamma+\varepsilon_n,\varepsilon_n\to 0$

Then easy to get $$\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{n+i}=\lim_{n\to\infty}\ln 2n-\ln n+\varepsilon_{2n}-\varepsilon_n=\ln 2$$