Determine the convergence of $\sum_{n=1}^{+\infty} \int_{0}^{\frac{1}{n}} \frac{\sin x}{1+x} dx$.
I'm working on determining the convergence of this number series.
$$\sum_{n=1}^{\infty} \int_{0}^{\frac{1}{n}} \frac{\sin x}{1+x} dx$$
It seems that it converges to $0.552$ using Mathematica, but I have no idea how to prove its convergence. Thanks in advance!
Solution 1:
$$\forall x\geq 0:~~ \frac{\sin(x)}{1+x}\leq x $$
Because, for all $x\geq 0$, $$\sin(x)\leq x,\quad \text{and}\quad \frac{1}{1+x}\leq 1$$
So $$\int_{0}^{\frac{1}{n}}\frac{\sin(x)}{1+x}dx\leq \int_{0}^{\frac{1}{n}}x \,dx = \frac{1}{2n^{2}}$$
And $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ is convergent.