Series Convergence

Calculus Level 5

$\large \sum_{n=1}^\infty \dfrac{(a-3\cos^2 n)\sin^2 n}{\sqrt n}$

Let $a$ be the constant value for which the series above converges.

If $a$ can be expressed as $\dfrac bc$ , where $b$ and $c$ are integers, with $b$ positive, find $b+c$ .

The answer is 7.

1 solution

Matija Sreckovic
May 7, 2016

Hint: Use Dirichlet's test.

First off, we need to use the trigonometric identities $sin^{2}n = \frac{1-cos2n}{2}$ and $cos^{2}n=\frac{1+cos2n}{2}$ a couple of times, and so we get that the $n$ th element of the series equals $\frac{4a-3-4acos2n+3cos4n}{8\sqrt{n}}$ .

Now, using Dirichlet's test, we can deduce that the series $\sum_{n=1}^{\infty} \frac{cos2n}{\sqrt{n}}$ and $\sum_{n=1}^{\infty} \frac{cos4n}{\sqrt{n}}$ converge (think complex sums!), and so the only way that our series will converge is if $4a-3=0$ (because the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges), which means $\boxed{a=\frac{3}{4}}$ .

If someone is unsure how to prove that $\sum_{k=1}^{n} cos2k$ and $\sum_{k=1}^{n} cos4k$ are bounded, ask me in the comments below, and I'll write it out.

Very nice problem and clear solution (+1)! Thanks!

To see that those partial sums are bounded, it may be worth noting that $\left|\sum_{k=1}^{n}\cos(2k\theta)\right|\leq|\csc(\theta)|$

Otto Bretscher - 5 years, 1 month ago

What i dont get, why 4a-3 has to be zero so that sum converge. In which case of it would not converge? 4a-3=1 will make it also converge, or actually any constant number.

Aleksandar Kovacevic - 5 years, 1 month ago

If, as you say, $a=1$ were to hold, then we'd get the series $\sum\limits_{n=1}^{\infty} \frac{1-4\cos2n+3\cos4n}{8\sqrt{n}}$ , which is the sum of two convergent series (the cosines) and one divergent series $\sum\limits_{n=1}^{\infty} \frac{1}{8\sqrt{n}}$ , and therefore must diverge.

The same goes for any other value $a\neq\frac{3}{4}$ . The sum of a divergent series and a convergent series is always divergent. The sum of two convergent series is convergent. The sum of two divergent series can both diverge and converge.

Matija Sreckovic - 5 years, 1 month ago

Wonderful approach...bows...(+1)!!!!

rajdeep brahma - 3 years ago

Nice problem and nice solution. Dirichlet test something I heard for the first time. It is interesting.

Srikanth Tupurani - 1 year, 10 months ago

Series Convergence

The answer is 7.

1 solution

0 pending reports