Convergence of this series: $\sum_{n=1}^{\infty}\frac{\sin(n)\sin(n^2)}{\sqrt{2n^2+1}}(x-1)^{2n}$?

Question

I have been trying for a long time to find an approach to this problem! I can't figure out how I could determine the convergence of this series!! I tried to determine the radius of convergence with the test of the report, but it seems that they do not succeed!

Answer 1

Welcome to MSE.

Note the series cannot converge if $|x-1|> 1$, i.e. $x<0$ or $x>2$, since then the terms fail the Test for Divergence. Convergence at the boundary is subtle, as always, so we'll deal with that later. To show the series converges on $(0,2)$, you could try the Absolute Comparison Test, which says: if $\sum_n |a_n|$ converges, then $\sum_{n} a_n$ converges. Then since $|\sin(\theta)|\leq 1$ for any real theta, we can throw them out: $$ \sum_{n=1}^{\infty}\left|\frac{\sin(n)\sin(n^2)}{\sqrt{2n^2+1}}(x-1)^{2n} \right| $$ $$ \leq \sum_{n=1}^{\infty}\left|\frac{1}{\sqrt{2n^2+1}}(x-1)^{2n} \right| = \sum_{n=1}^{\infty}\frac{1}{\sqrt{2n^2+1}}(x-1)^{2n} $$ For $0<x<2$, this series converges by the Ratio Test or by the Limit Comparison Test. At the boundary, $x=0$ and $x=2$, we have to examine the points individually. Here, plugging in the values gives the same result because of the even exponent, but that won't always be the case. The series in question is $$ x=0,x=2:\,\sum_{n=1}^{\infty}\frac{\sin(n)\sin(n^2)}{\sqrt{2n^2+1}}(1)^{2n}=\sum_{n=1}^{\infty}\frac{\sin(n)\sin(n^2)}{\sqrt{2n^2+1}} $$It turns out this series converges. You may have heard of the Alternating Series Test: if you have a series $S=\sum (-1)^n b_n$ such that:

• $b_n$ is decreasing
• $\lim\limits_{n\to\infty}b_n=0$;

then $S$ converges. There is a more general version known as Dirichlet's Test. It says: if you have a series $S=\sum a_n b_n$ such that:

• $b_n$ is decreasing
• $\lim\limits_{n\to\infty}b_n=0$
• there is a constant $M$ independent of $N$ such that $\displaystyle{\left|\sum_{n=1}^N a_n\right|\leq M}$;

then $S$ converges. We can take $b_n = \frac{1}{\sqrt{2n^2+1}}$; let's show if we have $a_n=\sin(n)\sin(n^2)$ that the last condition is satisfied. These partial sums have a closed form, which we can prove by induction.

Lemma: $\displaystyle{\sum_{n=1}^{N}\sin(n)\sin(n^2)=\sin^2\left(\frac{N(N+1)}{2}\right)}$. Proof: For $N=1$, this just says $\displaystyle{\sin^2(1)=\sin^2(1)}$, which is obvious. Suppose the result holds for some $N$: $$ \sum_{n=1}^{N}\sin(n)\sin(n^2)=\sin^2\left(\frac{N(N+1)}{2}\right) $$To go from $N$ to $N+1$, we just add the last term: $$ \sum_{n=1}^{N+1}\sin(n)\sin(n^2)=\sum_{n=1}^{N}\sin(n)\sin(n^2)+\left(\sin(N+1)\sin((N+1)^2)\right) $$ $$ =\sin\left(\frac{N(N+1)}{2}\right)\sin\left(\frac{N(N+1)}{2}\right)+\sin(N+1)\sin((N+1)^2) $$ $$ =\frac{1}{2} \left(1-\cos (N (N+1))+\cos \left(-(N+1)^2+N+1\right)-\cos \left((N+1)^2+N+1\right)\right) $$ $$ =\frac{1}{2} \left(1-\cos\left((N+1)(N+2)\right)\right) = \sin^2\left(\frac{(N+1)(N+2)}{2}\right), $$as we wanted to show. The point: this sum is clearly bounded by $1$, since it's $\sin^2(\text{something})$. Hence the series converges at the boundary by Dirichlet's Test, and therefore converges for $x\in[0,2]$.