Find limit with sum using integration

Question

I'm working on this problem:

Find the limit $$ \lim_{n \to \infty} \sum_{k=5n}^{7n} \frac{n}{k^2+n^2} $$

My initial thought is to turn it into an integral and work from there, but I'm not sure how to do this, especially since it contains $k=5n$.

Answer 1

Let us set

$$S_n=\sum_{j=5n}^{7n}\frac{n}{j^2+n^2}.$$

Observe that

$$S_n=\sum_{j=0}^{7n}\frac{n}{j^2+n^2}-\sum_{j=0}^{5n}\frac{n}{j^2+n^2}+\frac{n}{(5n)^2+n^2}=7\sum_{j=0}^{7n}\frac{7n}{(7j)^2+(7n)^2}-5\sum_{j=0}^{5n}\frac{5n}{(5j)^2+(5n)^2}+\frac{1}{26n}.$$

Set

$$a_{N,s}=\sum_{j=0}^N\frac{N}{(sj)^2+N^2}$$

so that

$$S_n=7a_{7n,7}-5a_{5n,5}+\frac{1}{26n}.$$

Observe now that

$$a_{N,s}=\frac{1}{s}\cdot\frac{s}{N}\sum_{j=0}^N\frac{1}{\left(\frac{s}{N}j\right)^2+1},$$

and so

$$\lim_{N\to\infty}a_{N,s}=\frac{1}{s}\int_0^s\frac{\mathrm{d}x}{x^2+1}=\frac{\arctan(s)}{s}.$$

It follows that

$$\lim_{n\to\infty}S_n=\arctan(7)-\arctan(5).$$

Answer 2

Your idea is fine, indeed we have that

$$\lim_{n\to \infty}\sum_{k=5n}^{7n} \frac{n}{k^2+n^2}=\lim_{n\to \infty}\frac1n\sum_{k=5n}^{7n} \frac{1}{\left(\frac kn\right)^2+1}=\int_5^7\frac1{x^2+1}dx=\arctan 7 - \arctan 5$$

Answer 3

Rewrite the limit as a difference of two limits $$ \lim _{n \rightarrow \infty} \sum_{k=5 n}^{7 n} \frac{n}{k^2+n^2}= \lim _{n \rightarrow \infty} \sum_{k=1}^{7 n} \frac{n}{k^2+n^2}-\lim _{n \rightarrow \infty} \sum_{k=1}^{5 n} \frac{n}{k^2+n^2}+ \underbrace{\lim _{n \rightarrow \infty} \frac{n}{(5 n)^2+n^2}}_{=0} $$ For the first limit, we have $$ \begin{aligned} \lim _{n \rightarrow \infty} \sum_{k=1}^{7 n} \frac{n}{k^2+n^2} = & \lim _{7 n \rightarrow \infty} \frac{7-0}{7 n} \sum_{k=1}^{7 n} \frac{1}{\left(0+\frac{(7-0)k}{7n}\right)^2+1}\\=& \int_0^7 \frac{1}{1+x^2} d x\\=&\arctan 7 \end{aligned} $$ Similarly for the second limit, we have $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{5 n} \frac{n}{k^2+n^2}=5\arctan 5 $$ We can now conclude that $$\boxed{\lim _{n \rightarrow \infty} \sum_{k=5 n}^{7 n} \frac{n}{k^2+n^2} = \arctan 7-\arctan 5 \;} $$

Replacing $5$ and $7$ by $a$ and $b$ respectively yields the general result $$\lim _{n \rightarrow \infty} \sum_{k=an}^{b n} \frac{n}{k^2+n^2}= \arctan b-\arctan a \; $$