# Thread: More on series converge or diverge?

1. ## More on series converge or diverge?

Hi again, as I was doing more questions on this, I got stuck on two more questions.

1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )
2. ∑(from n=1 to infinity) ( ln (1+1/n) )
Well Q1, same as before I tried ratio test and it turned out weird...
For Q2,not sure if I did it right, here's what I did. I compute the partial sums:
s1= ln(1+1)
s2= ln(1+1) + ln(1+1/2) > ln(1+1/2) + ln(1+1/2) = 2 ln(1+1/2)
and so lim for n to infinity for ln(1+1/n) = infinity
therefore the sequence of the partial sum diverges to infinity and so the sequence also diverges. Is this right?

Thanks.

2. Hello,

Originally Posted by Rui
Hi again, as I was doing more questions on this, I got stuck on two more questions.

1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )
You have to use the alternate series test, proving the absolute value of the summand is decreasing and that its limit when n tends to infinity is 0.
Or you can take a look at the alternate Riemann series, which state that $\displaystyle \sum \frac{(-1)^n}{n^a}$ converges if $\displaystyle a>0$

3. Originally Posted by Rui
1. ∑(from n=1 to infinity) ((-1)^(n+1)/ √n )
Use alternating series test:

$\displaystyle b_n = \frac 1{\sqrt{n}}$

$\displaystyle b_{n+1} = \frac 1{\sqrt{n+1}}$

$\displaystyle b_n \geq b_{n+1}$

$\displaystyle \lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac 1{\sqrt{n}} = 0$

So this satisfies the alternating series test, and is therefore convergent.
Originally Posted by Rui
2. ∑(from n=1 to infinity) ( ln (1+1/n) )
let $\displaystyle a_n = ln(1+1/n) ~~~~= ~~~~ln(\frac{n+1}n) ~~~~= ~~~~ln(n+1)-ln(n)$

Then lets evaluate this as a partial sum, looking at the telescoping aspect.

$\displaystyle \sum_{n=1}^k a_n$ $\displaystyle = ln(2)-ln(1)~~~~+~~~~ln(3)-ln(2)~~~~+~~~~ln(4)-ln(3)~~~~+~~~~...~~~~+~~~~ln(k)-ln(k-1)$

cancel out common terms, and see that -ln(1) = 0 to get
$\displaystyle \sum_{n=1}^k a_n = ln(k)$

so $\displaystyle \sum_{n=1}^\infty a_n = \lim_{k\to\infty} ln(k)$ which clearly diverges.