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The Lim 1/(n+r) Problem (TOMATO Subjective 155 with some modifications)

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


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.


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

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

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

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 

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


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