Limit of Riemann integrals

Question

Using Riemann integrals of suitably chosen functions, find

(a) $\lim_{n \to \infty} \displaystyle\sum_{k=1}^{2n} \dfrac{1} {n +k}$

(b) $\lim_{n \to \infty} \dfrac {1^k +2^k + ... + n^k}{n^{k+1}}$ for $k>0$

Answer 1

a) $$\lim_{n\to\infty}\sum_{k=1}^{2n}\frac{1}{n+k}=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1/n}{1+k/n}+\sum_{k=1}^{n}\frac{1/n}{2+k/n}\right)=\int_0^1\frac{dx}{1+x}+\int_0^1\frac{dx}{2+x}$$

Answer 2

a.) $$\lim_{n \to \infty} \displaystyle\sum_{k=1}^{2n} \dfrac{1} {n +k} = \lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{n+k}+\sum_{k=n+1}^{2n}\frac{1}{n+k}\right)$$ We then use the identity $\sum_{n=s+p}^{t+p} f(n-p) = \sum_{n=s}^{t} f(n)$ on the second summation to get $$\Rightarrow \lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{n}{n+k}+\sum_{k=1}^{n}\frac{n}{2n+k}\right)$$ $$ \Rightarrow \lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}+\sum_{k=1}^{n}\frac{1}{2+\frac{k}{n}}\right)$$ $$=\int_0^1 \frac{dx}{1+x} + \int_0^1 \frac{dx}{2+x}$$ $$ = \log(2) + \log\bigg(\frac{3}{2}\bigg) = \log(3)$$

b.) $$\lim_{n \to \infty} \dfrac {1^k +2^k + ... + n^k}{n^{k+1}}$$ $$ = \lim_{n \to \infty}\sum_{i=1}^{n} \frac{i^k}{n^{k+1}} = \lim_{n \to \infty}\sum_{i=1}^{n} \frac{i^k}{n^{k}}\frac{1}{n} = \lim_{n \to \infty}\sum_{i=1}^{n} \bigg(\frac{i}{n}\bigg)^k\frac{1}{n}$$ $$ = \int_{0}^{1} x^k dx$$