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}