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I have problem with proving that

$$a_{n} = - 2\sqrt {n+1} + \left({\frac{1}{\sqrt {1}}+{\frac{1}{\sqrt{2}}}+\cdots+{\frac{1}{\sqrt{n}}}}\right)$$

converges. I have tried to transform it to the $a_{n} = b_{n}+c_{n}$ where $c_{n}=-2{\sqrt{n+1}}$ and $b_{n} = \ldots$ but I'm not sure whether this way is correct. Thanks in advance!

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up vote 5 down vote accepted

Abel's partial summation technique: \begin{align*} \sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\ & = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\ & = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n)) \end{align*} (The above is nothing but the discrete version of integration by parts).

$$\sum_{n=1}^{N} a(n) f(n) = \int_{1^-}^{N^+} f(t) d(A(t)) = f(t) A(t) \rvert_{1^-}^{N^+} - \int_{1^-}^{N^+} A(t) f'(t) dt$$ (The second integral can be interpreted as a Riemann-Stieltjes integral.)

Consider the sum $\displaystyle \sum_{n \leq N} \frac1{\sqrt{n}}$. Choose $a(n) = 1$ and $f(n) = \frac1{\sqrt{n}}$. Note that we have $A(t) = \lfloor t \rfloor = t - \{t\}$. Hence, we get that \begin{align*} \sum_{n \leq N} \frac1{\sqrt{n}} & = \left. \frac{t-\{t\}}{\sqrt{t}} \right \rvert_{1^-}^{N^+} + \dfrac12\int_{1^-}^{N^+} \frac{(t-\{t\})}{t^{3/2}} dt\\ & = \sqrt{N} + \dfrac12\int_{1^-}^{N^+} \frac{dt}{t^{1/2}} - \dfrac12\int_{1^-}^{N^+} \frac{\{t\}}{t^{3/2}} dt\\ & = 2\sqrt{N} - 1 - \dfrac12\int_{1^-}^{N^+} \frac{\{t\}}{t^{3/2}} dt \end{align*} Hence, $$\sum_{n \leq N} \frac1{\sqrt{n}} - 2 \sqrt{N+1} = \underbrace{2\sqrt{N} - 2\sqrt{N+1}}_{\to 0 \text{(Why?)}} - \underbrace{\left(1 + \dfrac12\int_{1^-}^{N^+} \frac{\{t\}}{t^{3/2}} dt \right)}_{\to \text{constant (Why?)}}$$

With a little more effort you can show that $$\underbrace{-\left(1 + \dfrac12\int_{1^-}^{N^+} \frac{\{t\}}{t^{3/2}} dt \right) \to \zeta(1/2)}_{N \to \infty} \approx -1.46$$

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The limit is approximately $-1.46$ – Will Jagy Oct 25 '12 at 21:04
@WillJagy Thanks for pointing it out. I don't know what I had in mind to write it as $2$ instead of $\zeta(1/2)$. – user17762 Oct 25 '12 at 21:10

Maybe you want to approximate this sum by an integral:

$$ \int_1^{n} \frac{dx}{\sqrt{x}} = 2(\sqrt{n}-1)< \sum_{k = 1}^n \frac{1}{\sqrt{k}} < \int_0^{n-1} \frac{dx}{\sqrt{x}} = 2\sqrt{n-1} $$

For example:

$$ 2(\sqrt{16}-1)=6 < \frac{1}{\sqrt{1}} + \dots + \frac{1}{\sqrt{16}} \approx 6.664 < 2\sqrt{15}\approx 7.75$$

Then it's a little bit surprising, since both terms tend to infinity that:

$$ \lim_{n \to \infty}\sqrt{n+1}- \sqrt{n} = \lim_{n \to \infty}\frac{1}{\sqrt{n+1} + \sqrt{n}} = 0$$

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The easy way is to show that $$ a_{n} = - 2\sqrt {n+1} + ({\frac{1}{\sqrt {1}}+{\frac{1}{\sqrt{2}}}+\ldots+{\frac{1}{\sqrt{n}}}}) $$ begins a little low and increases with $n,$ while

$$ b_{n} = - 2\sqrt {n} + ({\frac{1}{\sqrt {1}}+{\frac{1}{\sqrt{2}}}+\ldots+{\frac{1}{\sqrt{n}}}}) $$ begins a little high and decreases with $n,$

finally $b_n > a_n$ and $b_n - a_n \rightarrow 0.$

The conclusion is that all the $b_j$ are larger than all the $a_i,$ so the $a_n$ is an increasing sequence with an upper bound.

Worth actually calculating the two and printing side by side for, say, $n \leq 50.$ I did that, it is not until $n \geq 159$ that I finally get $$ -1.5 < a_n < b_n < -1.4, $$ as in

159  -1.499903567256619  -1.420722711746567
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The conjugate formula $\sqrt{a}\pm\sqrt{b}=\frac{a-b}{\sqrt{a}\mp\sqrt{b}}$ is your friend here. Use it (twice) to show that $$ a_n-a_{n-1}=\frac1{\sqrt{n}}-2\sqrt{n+1}+2\sqrt{n}, $$ is $a_n-a_{n-1}=b_n$ with $$ b_n=\frac1{\sqrt{n}}-\frac{2}{\sqrt{n+1}+\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}(\sqrt{n}+\sqrt{n+1})}=\frac1{\sqrt{n}(\sqrt{n}+\sqrt{n+1})^2}. $$ Since $a_0=-2$, $a_n=-2+\sum\limits_{k=1}^nb_k$ hence the sequence $(a_n)_{n\geqslant0}$ increases from $a_0=-2$ to $\lim\limits_{n\to\infty}a_n=-2+\sum\limits_{k=1}^{+\infty}b_k$, where the last series converges since $0\leqslant b_n\lt\frac1{4n\sqrt{n}}$ for every $n\geqslant1$.

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