Question

INFINITE SUM
of sin(4n + 1)/4n^2
from n = 1 to infinity

As you can see here, that sum converges.:
http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+inf+of+sin(4n+%2B+1)+%2F+(4n%5E2)

I'm trying to show that it converges, but I can't think of which test to use.

If someone could at least tell me which test to use (without even having to tell me how to use it), I would very much appreciate it!

Answer 1

Comparison test should work. Hint: \(\sin x\) is bounded.

Answer 2

@SithsAndGiggles, thanks for your answer, but I preferred your first answer before you edited it. That instantaneously made me understand that the series converges, but I still cannot see how to link it to the direct comparison test ( http://en.wikipedia.org/wiki/Direct_comparison_test ), because it uses only one summation to compare.

In other words, I see that saying it's greater than a convergent series implies that the series is not diverging to negative infinity and that saying it's smaller than a convergent series implies that the series is not diverging to positive infinity, both of which together imply that it's converging, but the Wikipedia link I gave only compares to the series for which convergence has not been established with one series for which it has been established (instead of two like we did).

Answer 3

Could you please elaborate on this situation?

Answer 4

Sure. I edited my first response after rereading your question; you only wanted the test, but I didn't think it was enough of a hint.

The sine function is bounded:
\[-1\le\sin x\le1\]
The argument of the sine doesn't change the amplitude of the sine wave, which means \(\sin(\text{anything})\) will always be bounded between \(-1\) and \(1\), whereas \(k\sin(\text{anything})\) will be bounded between \(-k\) and \(k\). Hence
\[-1\le\sin(4n+1)\le1\]
Divide everything by \(n^2\) and you have
\[\frac{-1}{n^2}\le\frac{\sin(4n+1)}{n^2}\le\frac{1}{n^2}\]
which directly implies that
\[\sum_{n=1}^\infty\frac{-1}{n^2}\le\sum_{n=1}^\infty\frac{\sin(4n+1)}{n^2}\le\sum_{n=1}^\infty\frac{1}{n^2}\]
This is enough to show the given series converges. (I'm neglecting the 4 in the denominator because it doesn't change the conclusion.)

If you're not convinced, you can approach this a slightly different way by accepting that the numerator is bounded above by some constant \(c\) (which is actually 1). Then comparing to the series \(\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}\) will show you that the given series converges because
\[\sin(4n+1)\le c~~\iff~~\frac{\sin(4n+1)}{n^2}\le\frac{c}{n^2}\]
In other words, this "alternative" method ignores the lower bound of the numerator, since the upper bound is all that matters.