2
$\begingroup$

I need help in evaluating this problem: $\lim_{n \to \infty }\sum_{r=1}^{n}{\frac{1}{(n+r)(n+2r)}}$.

I tried to convert in to the form $\lim_{n \to \infty }\frac{1}{n}\sum_{r=1}^{n}{f\left(\frac{r}{n}\right)}$ to convert it into a definite integral, but it doesn't seem to be convertible to that form.

Is the answer $0$ because every term evaluates to zero if we apply the limit $ n \to \infty $?

The answer given in the book where I found it is $ln(1.5)$, but that answer will come if there's an $n$ in the numerator of every term. So, I'm not sure whether it is a misprint or there is some other method which I'm not aware of.

$\endgroup$
2
$\begingroup$

The answer is indeed zero. Let $$ f(x)=\frac{1}{(1+x)(1+2x)} $$ so that $\int_0^1 f(x)\, \mathrm{d}x=\ln(3/2)$. Now to prove that your limit is zero, use the rule $$ \lim_{n\to\infty}\left[f(x)\cdot g(x)\right]= \left(\lim_{n\to\infty} f(x)\right)\cdot\left(\lim_{n\to\infty} g(x)\right) $$ whenever both of the limits on the right-hand side exist:

$$ \lim=\lim_{n\to\infty}\left[\left(\frac{1}{n}\right)\cdot\left(\frac{1}{n}\sum_{r=1}^nf\left(\frac{r}{n}\right)\right)\right]=0\cdot\ln(3/2)=0. $$

If your book indicates that the answer is $\ln(3/2)$, then probably there is a misprint in either the problem or the answer. Unfortunately, misprints in textbooks (even math textbooks) happen all the time. Nicely spotted!

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

\begin{align} \sum_{r=1}^{n}{\frac{1}{(n+r)(n+2r)}} &<\sum_{r=1}^{n}\frac 1{n^2}\\ &=\frac 1n\\ &\xrightarrow{n\to\infty}0 \end{align}

On the other hand \begin{align} \sum_{r=1}^{n}{\frac{n}{(n+r)(n+2r)}} &=\sum_{r=1}^{n}\frac 1{(1+\frac rn)(1+2\frac rn)}\frac 1n\\ &=\int_0^1\frac{\mathrm dx}{(1+x)(1+2x)}\\ &=\log\left(\frac 32\right) \end{align} hence there is a typo in your book.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.