How to evaluate the limit of this following expression $\color{black}{\lim_{n\to \infty}\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}}$?

Question

I am not able to find the limit of the following expression, after trying a bit hard:$$\lim_{n\to \infty}\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}$$ Can some one help with some approach?

Use: Definite Integral and apply the limit of sum formula => https://prnt.sc/vu3x6i — Jon Snow, Dec 02 '20 at 07:43
@ jon I don't have that idea I only know upto L hospital's rule, that is why may be I am not able to do it! Can you throw some more light , i.e how to do such things? — random, Dec 02 '20 at 07:46
Look at @Archis Welankar answer. He went a bit more in-depth / almost to the answer. — Jon Snow, Dec 02 '20 at 07:47

score 2 · Answer 1 · answered Dec 02 '20 at 07:45

Taking $n^2$ common from denominator we have $\frac{6n}{9n^2-r^2}=\frac{6}{9-\frac{r^2}{n^2}}\frac{1}{n}$ . Now observing the fact that it's a Riemann sum we have $\lim_{n\to \infty} \sum_1^{n} \frac{6n}{9n^2-r^2}=\int _0^1 \frac{6}{9-x^2}dx$ . I hope you can continue from here.

grand_chat · Answer 2 · 2020-12-02T19:25:34.680

Seeing as this question is tagged as precalculus and contest math, here's a solution that avoids Riemann sums:

Write $$ \frac{6n}{9n^2-r^2}=\frac{6n}{(3n+r)(3n-r)}=\frac1{3n+r}+\frac1{3n-r} $$ and sum over $r=1,\ldots,n$ to find that $$ \frac1{3n} + \sum_{r=1}^n \frac{6n}{9n^2-r^2}=\frac 1{2n} + \frac1{2n+1} +\cdots + \frac1{4n-1}+\frac1{4n}. $$ If you're familiar with sums of the form $S_n:=\frac1{an}+\cdots+\frac1{bn}$ (for $b>a$) you know that the RHS converges to $\log (b/a)=\log 2$.

For completeness, here's an elementary derivation.

Claim: Assume $b>a$. Then $$\lim_{n\to\infty} \left(\frac 1{an}+\cdots+\frac1{bn}\right) = \log\left(\frac ba\right).$$

Proof: Take the inequality $$ \frac{x-1}x\le \log x \le x-1 $$ and substitute $x=1+\frac1k$. This yields for every $k$ $$ \frac1{k+1}\le \log(k+1)-\log k\le \frac 1k. $$ Sum from $k=an$ to $bn$ and use telescoping to get $$S_n-\frac1{an}+\frac1{bn+1}\le \log(bn+1)-\log(an)\le S_n. $$ Rearrange this to the form $$ \log\left(\frac{bn+1}{an}\right)\le S_n\le \log\left(\frac{bn+1}{an}\right)+\frac1{an}-\frac1{bn+1}. $$ Take the limit as $n\to\infty$ and apply the squeeze theorem to obtain the result.

score 1 · Answer 3 · edited Dec 02 '20 at 08:19

1

Elaborating on @Archis Welankar's answer:

Once you've solved the Limit of Sum definition of the series - limit, you then simply solve for the integral.

$ \int _0^1 \frac{6}{9-x^2}dx = \mathbf{\ln(2)}$

edited Dec 02 '20 at 08:19

J.G.

115,835

answered Dec 02 '20 at 07:52

Jon Snow

58

I've done my best to honour your make-the-answer-bold intent within MathJax. – J.G. Dec 02 '20 at 08:19
Thanks, I appreciate it! <3 – Jon Snow Dec 02 '20 at 08:47

score 1 · Answer 4 · answered Dec 02 '20 at 08:34

We can even compute the partial sums $$S_n=\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}=\sum_{r=1}^n\frac{1}{3 n+r}+\sum_{r=1}^n\frac{1}{3 n-r}$$ $$S_n=\big[\psi (4 n+1)-\psi (3 n+1) \big]+\big[\psi (1-3 n)-\psi (1-2 n)\big]$$ Using asymptotics $$S_n=\log (2)+\frac{1}{24 n}+\frac{1}{64 n^2}+O\left(\frac{1}{n^4}\right)$$

How to evaluate the limit of this following expression $\color{black}{\lim_{n\to \infty}\sum_{r=1}^n\frac{6n}{9{n}^2-{r}^2}}$?

4 Answers4