Series is convergent but it seems it is divergent?

I have a series:

$$\sum^\infty_{n=1}{\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}\bigg)}$$

and I thought it is a divergent series since

$$\sum{\big(f(x)-g(x)\big)} = \sum{f(x)} - \sum{g(x)}$$

and so the series equals to

$$=\sum^\infty_{n=1}{\frac{1}{\sqrt{n}}}-\sum^\infty_{n=1}{\frac{1}{\sqrt{n+2}}}$$

and we know that $\sum^\infty_{n=1}{\frac{1}{\sqrt{n}}}$ is divergent, so the whole series is divergent. But it turns out that it is convergent and the answer is $1+\frac{1}{\sqrt{2}}$.

How do you prove that it is convergent and calculate the answer? The only way I know how to compute an answer for a series is via a geometric series and there seems to be no way to make it into a geometric series!

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@labbhattacharjee - Oh wow thank you so much! I totally forgot telescoping series existed! –  Derek 朕會功夫 Apr 4 '14 at 5:35
Consider $\sum_{i=1}^\infty 1$, which diverges, and $\sum_{i=1}^\infty 1$, which diverges. But $\sum_{i=1}^\infty (1-1)$ obviously converges to 0. –  MJD Apr 4 '14 at 5:43

The rule you quoted above $\sum (f(x)-g(x)) = \sum f(x) - \sum g(x)$ is not correct if one or both of the series on the right is divergent. In this case, both $\sum \frac{1}{\sqrt{n}}$ and $\sum \frac{1}{\sqrt{n+2}}$ are divergent.

As indicated in the comments, to evaluate this series, you should telescope the series.

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To further elaborate as to why we cannot say $\sum f(x) - g(x)$ diverges if $\sum f(x)$ and $\sum g(x)$ are individually divergent, consider the example $$\sum_{k=1}^\infty \frac{1}{k(k+1)} < \sum_{k=1}^\infty \frac{1}{k^2} < \infty.$$ But we can of course write $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1},$$ and we know that $$\sum_{k=1}^\infty \frac{1}{k}$$ is divergent. Indeed, even more trivially, $$\sum_{k=1}^\infty 0 = 0,$$ yet we could write it as $$\sum_{k=1}^\infty (1-1) = \sum_{k=1}^\infty 1 - \sum_{k=1}^\infty 1 = \infty - \infty.$$ –  heropup Apr 4 '14 at 6:08
The summation rule is valid if both of the individual series converge. Just check the partial sums. –  Ryan Reich Apr 4 '14 at 15:09
@RyanReich Yes you're right. –  Ted Apr 4 '14 at 15:43
@heropup Those are a nice set of examples; I think you should post that comment as an answer. –  Mike Miller Apr 5 '14 at 0:30

Hint

Write $$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}=\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\bigg)+\bigg(\frac{1}{\sqrt{n+1}}-\frac{1}{\sqrt{n+2}}\bigg)$$ and then telescope.

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$$\sum_{n=1}^\infty \bigg(f(x)-g(x)\bigg)=\sum_{n=1}^\infty f(x)-\sum_{n=1}^\infty g(x)$$

is valid only when both $f$ and $g$ are convergent. An infinite sum is nothing more than the limit of partial sums, and the above is equivalent to writing,

$$\lim_{n\to\infty} (S_{n_1}-S_{n_2}) = \lim_{n\to\infty} S_{n_1} - \lim_{n\to\infty}S_{n_2}$$

which is only right when both the individual limits exist.

Now moving on to the problem at hand,

$$\sum_{n=1}^\infty \frac1{\sqrt{n}} -\frac1{\sqrt{n+2}}$$

$$S_n=1-\frac1{\sqrt{3}}+\frac1{\sqrt{2}}-\frac1{\sqrt{4}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\cdots +\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}$$

$$S_n=1+\frac1{\sqrt{2}}-\frac{1}{\sqrt{n+2}}$$

Clearly, we have,

$$\lim_{n\to\infty} S_n=1+\frac{1}{\sqrt{2}}$$

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I was wondering why the answer was $1$... and then I saw the edit. –  Derek 朕會功夫 Apr 4 '14 at 5:46
@Derek朕會功夫 yeah typo. –  Sabyasachi Apr 4 '14 at 5:54

Note: $$\sum_{n = 1}^\infty \left[\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}\right] = 1 - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{6}} + \cdots = 1 + \frac{1}{\sqrt{2}}.$$

By the way, if $\sum f(n)$ and $\sum g(n)$ diverge, it does not mean $\sum [f(n) - g(n)]$ diverges because, for example, $$\sum_{n = 1}^\infty \left[\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}\right]$$ converges.

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"The rule you are trying to use to disprove X is incorrect; for a counterexample, observe that X is true". I'm not sure whether that's brilliant or horrible. –  Rawling Apr 4 '14 at 11:49
@Rawling If there is a counterexample, then the "theorem" is obviously not true. –  glebovg Apr 4 '14 at 20:06
Well yes, but if he's trying to use it to disprove X, X isn't going to be the first counterexample that springs to mind :) –  Rawling Apr 5 '14 at 7:03
@Rawling I see what you mean. –  glebovg Apr 5 '14 at 10:22