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I have the sequence $a_n =\sqrt{n+1}-\sqrt{n}$ and $b_n=\dfrac{1}{\sqrt{n}}$

let $s_n$=$a_1 + a_2 + a_3 + ... + a_n =\displaystyle \sum_{k=1}^n a_k$

and $t_n$=$b_1 + b_2 + b_3 + ... + b_n = \displaystyle \sum_{k=1}^n b_k$

be the corresponding sequences of partial sums.

How would i find $s_5, s_{20}$ and $s_{100}$?

I then need to simplify and show that $s_n$ tends to infinity

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  • $\begingroup$ (exact-sequence) is something from algebra, or not ? $\endgroup$ – inequal Jan 16 '15 at 12:23
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Notice that $s_n$ is a telescopic sum:

$$s_n=\sum_{k=1}^na_k=\sum_{k=1}^n(\sqrt{k+1}-\sqrt k)=\sqrt{n+1}-\sqrt 1=b_{n+1}-b_1$$ and we can also find the same result by changing the index $$s_n=\sum_{k=1}^n a_k=\sum_{k=1}^n( b_{k+1}-b_k)=\sum_{k=1}^nb_{k+1}-\sum_{k=1}^nb_k=\sum_{k=2}^{n+1}b_k-\sum_{k=1}^nb_k=b_{n+1}-b_1$$

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Hint: $s_n = \sqrt{n+1} - 1$. Can you continue??

Also you can find out some thing about $t_n$'s convergence status by using:

$b_n > 2\left(\sqrt{n+1} - \sqrt{n}\right) = 2a_n$

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