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Problem:  Evaluate: $\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n})$

Solution:

As the title suggests the modification of this problem will be, that we will solve a more general series and then use a specific value to arrive at the solution of this problem.

First let us consider the following limit:

$\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+kn})$

Observe carefully that using k=1 in this limit, we get the limit that has been asked to evaluate.

Now

$\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}) = \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{1}{n+r})$

$= \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{\frac{1}{n}}{1+\frac{r}{n}})$

$= \lim_{n\to\infty}\frac{1}{n} (\sum_{r=1}^{kn} \frac{1}{1+\frac{r}{n}})$

Let’s substitute $\frac{r}{n} = x => dr = ndx$

Now we can change the summand to an integral

$=> \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{\frac{1}{n}}{1+\frac{r}{n}}) = \lim_{n\to\infty} \frac{1}{n}*n\int_{0}^{k} \frac{1}{1+x} dx$

$= \lim_{n\to\infty}( log |x+1|_{k} - log |x+1|_{0})$

$= \lim_{n\to\infty} log |k+1|$

$= log |k+1|$                     (As the term is an ‘n’ free term)

So we see the solution is $= log |k+1|$

Substituting k=1, we get

$\lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}) = \log {2}$