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Have two series, just a quick check of some simple series:

$\sum _{1}^{\infty} \frac {1}{\sqrt {2n^{2}-3}}$

Considering $\frac {1}{\sqrt {2n^{2}-3}}$ > $\frac {1}{\sqrt {4n^{2}}}$ = $\frac {1}{2n}$

Since $\sum _{1}^{\infty} \frac {1}{2n}$ $\rightarrow$ Diverges, hence by the camparsion test we have that $\sum _{1}^{\infty} \frac {1}{\sqrt {2n^{2}-3}}$ diverges.

The second one is $\sum _{1}^{\infty} (-1)^{n}(1+\frac{1}{n})^{n}$

Consdering $(-1)^{n}(1+\frac{1}{n})^{n} \le |(-1)^{n}(1+\frac{1}{n})^{n}|= (1+\frac{1}{n})^{n}$

We know that $(1+\frac{1}{n})^{n}$ converges, hence we have that $\sum _{1}^{\infty}(1+\frac{1}{n})^{n}$ coverges so again by the comparison test we have that $\sum _{1}^{\infty} (-1)^{n}(1+\frac{1}{n})^{n}$ converges absolutely $\Rightarrow$ converges,

many thanks in advance.

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$\frac{1}{n}$ converges, but $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges... –  Aryabhata Mar 1 '12 at 20:14
    
First one is OK, but start at $n=2$, or your calculator will blow up. The second is very not OK, your terms are alternately close to $e$ or close to $-e$, so partial sums bounce around a lot, and don't settle down. –  André Nicolas Mar 1 '12 at 20:20
    
So would i just show that the sequence converges too two different limits, say for example $a_{2n+1}$ and $a_{2n}$. –  user24930 Mar 1 '12 at 20:47

2 Answers 2

up vote 1 down vote accepted

The first series you mention diverges, and the reason you give is correct. However, it is not correct to say "$\to$ Diverges", as this would be as saying "approaches diverges" (or "converges to diverges"), which makes no sense. You could say that it "diverges to infinity" or write $\to +\infty$. Anyway, this is minor. Your argument is fine.

On the other hand, the second series diverges, and your argument is incorrect. This is because the sequence $(1+\frac1n)^n$ converges, but that is not what is being asked. You are asked to add up together the terms of the sequence, which is a completely different thing.

The usual way of dealing with the second example is to use the $n$-th term test, which says that if a series converges, its $n$th term converges to 0. Here, the terms are $(-1)^n\left(1+\cfrac1n\right)^n$ which do not converge to 0 because they do not converge at all. In fact, in absolute value, they approach $e$. It follows that the series diverges.


Note that you do not need to know or recognize that the terms of the second series in absolute value approach $e$ to conclude that the series diverges. Just note that $$ \left(1+\frac1n\right)^n\ge 1, $$ so also $|(-1)^n(1+\frac1n)^n|\ge1$, and the terms cannot converge to 0.

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Oh is this the non-null test, if so totally forgot about that. –  user24930 Mar 1 '12 at 21:00
    
Yes, that's it. I had seen it called that way in a while. –  Bruce George Mar 1 '12 at 21:05

The first argument is ok (you should start the series at $n=2$ though).

The second is not. As you imply, for the second series, you should be able to compute the value of the limit of (part of) the sequence of terms you are adding: $\lim\limits_{n\rightarrow\infty}(1+{1\over n})^n$. Then, based on the value of this limit, can the series converge?

Hint: The limit above is not zero. Thus $\lim\limits_{n\rightarrow\infty}[(-1)^n(1+{1\over n})^n]\ne0$. What can you say about a series $\sum\limits_{n=1}^\infty a_n$ if $\lim\limits_{n\rightarrow\infty}a_n\ne0$?

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Many thanks for that, it diverges due to the non-null test? –  user24930 Mar 1 '12 at 21:02
    
@user24930 If "non-null test" means what I think it means, then yes. –  David Mitra Mar 1 '12 at 21:11

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