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$$(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…$$

I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent.

Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges?

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By telescoping, we see that the $n$th partial sum is $\sqrt {n+1}-1$, so this series diverges.

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    $\begingroup$ This also answers the part of the question about the sum being a natural number when the number of terms is one less than a square. $\endgroup$ – robjohn Oct 7 '14 at 14:53
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So your series can be written as $$\sum_{n=1}^{\infty} (\sqrt{n+1}-\sqrt{n}) =\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} \geq \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}}$$ and so by Comparison Test diverges

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The $n$ term may be rewritten as

$$\frac1{\sqrt{n}+\sqrt{n+1}}$$

which behaves as $1/(2 \sqrt{n})$ as $n \to\infty$, so by comparison with the harmonic series, this series diverges.

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@Jasper Joy is indeed right. It's a telescoping series whose partial sums eventually only have a fixed number of terms after cancellation:

$$(\sqrt{n+1} - \sqrt n) + ... + (\sqrt 4 - \sqrt 3) + (\sqrt 3 - \sqrt 2) + (\sqrt 2 - \sqrt 1)$$ $$= \sqrt{n+1} + (-\sqrt n + \sqrt n) + ... + (-\sqrt 3 + \sqrt 3) + (-\sqrt 2 + \sqrt 2) - \sqrt 1$$ $$= \sqrt{n+1} - 1$$

In series sigma notation we can summarize this as:

$$\sum_{i=1}^{n} (\sqrt{i+1}-\sqrt{i}) = {\sqrt{n+1} - 1}$$

As an infite sum it diverges to $\infty$:

$$\sum_{i=1}^{\infty} (\sqrt{i+1}-\sqrt{i}) = \lim_{i \to \infty}{\sqrt{i+1} - 1}$$

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    $\begingroup$ We know @Jasper is right. Why duplicate? $\endgroup$ – Did Oct 7 '14 at 19:41
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    $\begingroup$ So those who are unfamiliar with telescoping can see the details of how it is done. $\endgroup$ – soakley Oct 7 '14 at 22:03

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