1

I am trying to figure out whether this converges absolutely or conditionally.

$ \sum^{\infty}_{n=4} \frac{(3)(-1)^n}{\sqrt{n}+3log(n)} $

Hence, I would like to figure out whether the absolute value or $ 3 \sum^{\infty}_{n=4} \frac{1}{\sqrt{n}+3log(n)} $ converges.

I thought to try to tackle this with Gauchy condensation:

$ 3 \sum^{\infty}_{n=4} \frac{2^n}{{2^{(1/2)n}}+3log(2^n)} $ = $ 3 \sum^{\infty}_{n=4} \frac{2^n}{{2^{(1/2)n}}+n3log(2)} $.

From here, it appears to diverge since the numerator grows much fastor than the denominator. Correct?

Haim
  • 757

2 Answers2

1

The series converges conditionally but not absolutely. It converges conditionally because for large $m$, when $n > m$ we have $f(n) = \frac{3}{\sqrt{n} + 3 \log n} \to 0$ monotonically as $n \to \infty$ and thus by the Alternating series test it converges.

The series $\sum_{n \geq 4} \frac{1}{\sqrt{n} + 3 \log n}$ diverges. This series is a positive series. Note that $\lim_{n \to \infty} \frac{\log n}{\sqrt{n}} = 0$. We have \begin{align} \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n} + 3 \log n}} = \lim_{n \to \infty} \frac{\sqrt{n} + 3 \log n}{\sqrt{n}} = 1 + 3\lim_{n \to \infty} \frac{\log n}{\sqrt{n}} = 1 \end{align} with $\sum_{n \geq 4} \frac{1}{\sqrt{4}}$ diverging. Thus by the limit comparison test he series $\sum_{n \geq 4} \frac{1}{\sqrt{n} + 3 \log n}$ diverges.

Jeevan Devaranjan
  • 2,575
0

No need to use the Cauchy Condensation Test.

I showed in THIS ANSWER, that $\log(n)\le \sqrt{n}$ for $n\ge 1$.

Then, note that for $n\ge 1$,

$$a_n=\frac{3}{\sqrt{n}+3\log(n)}\ge \frac{3}{4\sqrt{n}}$$

Finally, using the comparison test we see that the series of interest does not absolutely converge.

To test for conditional convergence, we note that $a_{n+1}\le a_n$ for all $n\ge 1$. Hence, $a_n$ monotonically decreases. Leibniz's Rule for alternating series guarantees that the series $\sum_{n=1}^\infty \frac{3(-1)^n}{\sqrt{n}+3\log(n)}$ converges.

Mark Viola
  • 179,405