# Convergence of alternating series

1. Jan 20, 2016

### Incand

1. The problem statement, all variables and given/known data
Do the following series converge or diverge?
$\sum_{n=2}^\infty \frac{1}{\sqrt{n} +(-1)^nn}$ and
$\sum_{n=2}^\infty \frac{1}{1+(-1)^n\sqrt{n}}$.

2. Relevant equations
Leibniz convergence criteria:
If $\{a_n\}_{k=1}^\infty$ is positive, decreasing and $a_n \to 0$, the alternating series $\sum_{n=1}^\infty (-1)^{n-1}a_n$ is convergent.

3. The attempt at a solution
I suspect the first series converges and the second diverges but I need to show that.
Starting with the first series it can be rewritten
$\sum_{n=2}^\infty \frac{1}{\sqrt{n}}\frac{(-1)^n}{(-1)^n+\sqrt{n}}$
At this point I had hopes that there exists an $N$ for which $\forall n \ge N$ $a_n$ is decreasing. That is show that
$\frac{1}{\sqrt{n+1}}\frac{1}{-1+\sqrt{n+1}} \le \frac{1}{\sqrt{n}}\frac{1}{1+\sqrt{n}}$. Sadly based on some numerical experiments this doesn't seem to be true so I need another approach.

Another idea was to be able to pair together the coefficients
$\frac{1}{\sqrt{n+1}}\frac{1}{-1+\sqrt{n+1}}+ \frac{1}{\sqrt{n}}\frac{1}{1+\sqrt{n}}$ to show that the partial sum $\lim_{n\to \infty} S_{2n}$ exists but I see no useful way to show this either.

2. Jan 20, 2016

### SammyS

Staff Emeritus
You should be able to show the following.

$\displaystyle \frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})} \ge \frac{1}{\sqrt{n}}\frac{1}{(1+\sqrt{n})}$

then ...

Does $\displaystyle \lim_{{n\to\infty}} \left(\frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})}\right) = 0 \ ?$

3. Jan 21, 2016

### Incand

Both of these statements are obviously true (the first since $(-1+\sqrt{n}) < (1+\sqrt{n})$). But I don't see how either one helps me ?
I still have that alternating sign that doesn't allow me to use the comparison test. For example I can't say that $\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}((-1)^n+\sqrt{n})} \le \sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}(-1+\sqrt{n})}$and neither series converges absolutely either.

I'm not sure I should use the Leibniz criteria either, these exercises are taken from a section covering that but I suspect they're meant to show cases where I can't apply that criteria directly or perhaps not at all and have to find other methods.

4. Jan 21, 2016

### Incand

I think I solved it now. Summing up the $2n$ and $2n+1$ terms seem to work.
$\frac{1}{\sqrt{2n}+2n}+ \frac{1}{\sqrt{2n+1}-2n-1} = \frac{\sqrt{2n+1}+\sqrt{2n}-1}{-4n^2+\dots}$
which is of order $\frac{1}{n^{3/2}}$ and hence converges.

While for the second one doing the same thing
$\frac{1}{1+\sqrt{2n}} + \frac{1}{1-\sqrt{2n+1}} = \frac{\sqrt{2n}-\sqrt{2n+1}+2}{2n}$ which is of order $\frac{1}{\sqrt{n}}$ and hence diverges if you do compare it against the harmonic series.

5. Jan 22, 2016

### SammyS

Staff Emeritus
What I intended with my suggestion was to suggest a way for you to use the Leibniz criteria.

You have for the first series: $\displaystyle \ \sum_{n=2}^\infty \frac{1}{\sqrt{n}}\frac{(-1)^n}{(-1)^n+\sqrt{n}} = \sum_{n=2}^\infty (-1)^n a_n \$

Thus $\displaystyle \ a_n = \frac{1}{\sqrt{n}}\frac{1}{(-1)^n+\sqrt{n}} \ .$ We see that an is positive. for all n ≥ 2, so the series is indeed alternating.

To see that the sequence, [ an ] converges, compare it to the sequence, [ bn ], where
$\displaystyle b_n = \frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})} \ .$​

When n is odd, an = bn .

When n is even, an < bn , which is consistent with the first inequality I gave in post #2, and is:
$\displaystyle \frac{1}{\sqrt{n}}\frac{1}{(1+\sqrt{n})}<\frac{1}{\sqrt{n}}\frac{1}{(-1+\sqrt{n})}$​

I think you'll find that the sequence, [bn] converges to 0, therefore, [an] converges to zero

6. Jan 22, 2016

### Incand

Right that's a good way to show that $a_n$ converges. I never thought about that part since it seemed quite obvious that the sequence converges with the $\sqrt{n}$ factor in the denominator. But this still wouldn't allow us to use the Leibniz criteria right since we don't have a decreasing sequence?

So let's see if I understand this right, your post would be a way to start if Leibniz criteria would be applicable, but in this case it turned out to not be? I included the criteria in the post since most problem in the same section seemed to be using that criteria so I thought it would be of use.

7. Jan 22, 2016

### SammyS

Staff Emeritus
Oh! Of course.

I missed the word 'decreasing'.

DUH !

8. Jan 22, 2016

### Incand

Then we're on the same page! It didn't help that I thought I should use that criteria for a problem it didn't apply to at the start.