# Infinite series convergence question

$$\sum_{n=3}^{\infty}\frac{(-1)^n}{\log n}$$

Can the conditional convergence of this series be proved by alternating series test, since you need n to be a natural number for the alternating series test and here we have n is a natural number from 3 so i dont know if the condition is met

• Specifically, what part of applying the alternating test are you having difficulties with? – Simply Beautiful Art Dec 3 '17 at 13:50
• It says that one of the conditions of the alternating test is that n has to be any natural number but here ive got n is bigger than or equal to 3 so is the condition still met so we can use the alternating test? – user504498 Dec 3 '17 at 13:55
• Yes, that condition is met. – Simply Beautiful Art Dec 3 '17 at 13:57
• Alternatively, you could rewrite $$\sum_{n=3}^\infty{(-1)^n\over\log(n)}=\sum_{n=0}^\infty{(-1)^{n+1}\over\log(n+3)}$$ – Simply Beautiful Art Dec 3 '17 at 13:59

Two points -

2) For convergence of the series can we apply Alternating series test?

Well if in the series $\sum_{n=1}^{\infty}(-1)^{n}a_{n}$ if $a_{n}$ is a monotonically decreasing sequence and also $\lim_{n \rightarrow \infty} \frac{1}{\log(n)} \rightarrow 0$ so we can apply!

So is $\frac{1}{\log(n)}$ decreasing? Yes it tends to zero as $n \rightarrow \infty$,thus the series is convergent.

Well if $n$ indexing starts from $n=1$ then we have the first term $-\frac{1}{\log(1)}$ which is infinity so it diverges right!.

So if your question would have been $\sum_{n \in \Bbb{N}} (-1)^n \frac{1}{\log(n)}$ then the series would diverge!

I think you are confusing with the Indexing!

Observe that in series $\sum_{n=1}^{\infty} (-1)^{n} a_{n} = \sum_{n=k}^{\infty} (-1)^{n-k} a_{n-k}$!

So applying to your question $\sum_{n=3}^{\infty} (-1)^{n} \frac{1}{\log(n)} = \sum_{n=0}^{\infty} (-1)^{n+3} \frac{1}{\log(n+3)}$ and you can now apply the ratio test as above!

• The series in question starts at $n=3$. – Simply Beautiful Art Dec 3 '17 at 13:59
• How do i prove its only conditional convergence and not absolute what should i add to what u wrote above – user504498 Dec 3 '17 at 14:02
• @SimplyBeautifulArt perhaps it is clear now? – BAYMAX Dec 3 '17 at 14:04
• @user504498 if you are talking about absolute convergence then you are dealing with $\sum_{n=3} \frac{1}{\log(n)}$,now $\sum_{n=3} \frac{1}{\log(n)} > \sum_{n=3}^{\infty} \frac{1}{n}$ so it diverges! much more here – BAYMAX Dec 3 '17 at 14:08

We can rewrite $$\sum_{n=3}^{\infty}\frac{(-1)^n}{\log n}=\sum_{n=3}^{\infty}(-1)^n \frac{1}{\log n}$$ which is in the form of $(-1)^nb_n$, which means that we can use the alternating series test. As $\frac{1}{\log n}$ never is infinity, it decreases and $$\lim_{x\to\infty}\frac{1}{\log n}=0$$ it converges.