How to prove $f(n)=1+1/\sqrt2+1/\sqrt3+\cdots +1/\sqrt n-2\sqrt n$ is convergent?

Question

I cant find any clue, but I hope if I am able to show the sequence is monotone decreasing and bdd below then we are done. I can see the sequence us monotone decreasing, the only disturbing part which I can't is how to bdd below. I've also checked that $f(1)=-1$, $f(2)>-1$ and so on. But I'm unable to show for all $n$. Please help me to solve this. Thanks in advance.

Answer 1

Note that

\begin{align*} f(n) &= \sum_{k=1}^{n} \left( \frac{1}{\sqrt{k}} - 2(\sqrt{k} - \sqrt{k-1}) \right) \\ &= \sum_{k=1}^{n} \left( \frac{1}{\sqrt{k}} - \frac{2}{\sqrt{k} + \sqrt{k-1}} \right) \\ &= \sum_{k=1}^{n} \frac{\sqrt{k-1} - \sqrt{k}}{\sqrt{k}(\sqrt{k} + \sqrt{k-1})} \\ &= - \sum_{k=1}^{n} \frac{1}{\sqrt{k}(\sqrt{k} + \sqrt{k-1})^2} \end{align*}

I hope this is enough for you to conclude the convergence of $f(n)$. Also, if we are allowed to use a bit of calculus, then we can come up with a more systematic solution rather than this tricky approach. (Which will be essentially a toy version of Euler-MacLaurin expansion.)

Answer 2

HINT.-$f(x+h)-f(x)=2\sqrt x+\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{x+2}}+\cdots+\dfrac{1}{\sqrt{x+h}}-2\sqrt{x+h}$.

It is easy enough to prove that $\lim_{x\to\infty}(2\sqrt x-2\sqrt {x+h})=0$.

Then $\{f(n)\}_{n\in\mathbb N}$ is a Cauchy sequence so it have a limit in $\mathbb R$.

Answer 3

Your expression is bounded below because it is decreasing, while $$g(n)=1+1/\sqrt2+1/\sqrt3+\cdots +1/\sqrt n-2\sqrt { n +1 }$$ is smaller and increasing. You must prove the increasing and decreasing claims, of course.

The limit of both sequences is roughly $\; \; -1.46035 \; \; \; $ You can see this value early by printing $$h(n)=1+1/\sqrt2+1/\sqrt3+\cdots +1/\sqrt n-2\sqrt { n + \frac{1}{2} }$$ which has a much smaller "tail" than the others. For the same reason, I'm not sure whether $h$ increases or decreases; at least, I don't remember

   n       sum              sum - 2 sqrt(n)    sum - 2 sqrt(n+1)
   1   1.0                -1.0                -1.82842712474619
   2   1.707106781186547  -1.121320343559643  -1.756994833951207
   3   2.284457050376173  -1.179644564761581  -1.715542949623827
   4   2.784457050376173  -1.215542949623827  -1.687678904623406
   5   3.231670645876131  -1.240465309123449  -1.667308839690225
   6   3.639918936339994  -1.259060549226362  -1.651583685789187
   7   4.017883409349222  -1.27361921277996   -1.638970840143159
   8   4.371436799942495  -1.285417449549885  -1.628563200057505
   9   4.704770133275828  -1.295229866724172  -1.619785187060931
  10   5.020997899292666  -1.303557421044093  -1.612251681418133
  11   5.32250924387043   -1.31074033684037   -1.605693986405079
  12   5.611184378465243  -1.317018851810266  -1.599918172462735
  13   5.888534476577857  -1.322568074350121  -1.594780296970026
  14   6.155795718490282  -1.327519055057601  -1.590170973924552
  15   6.413994608237443  -1.331972084177391  -1.586005391762557
  16   6.663994608237443  -1.336005391762557  -1.582216642997878
  17   6.906530233273776  -1.339681017961545  -1.578751140964793
  18   7.142232493669292  -1.343048880569278  -1.575565393412056
  19   7.371648227539854  -1.346149659541494  -1.572623682459305
  20   7.595255025289833  -1.349016884709326  -1.569896364621846
  21   7.813472915525826  -1.351678474385854  -1.567358604121034
  22   8.026673631881437  -1.354157887765423  -1.564989414744002
  23   8.235188045938511  -1.356475000686928  -1.562770925194201
   n       sum              sum - 2 sqrt(n)    sum - 2 sqrt(n+1)

Answer 4

Approach $\bf{1}$

One way to bound such series is to note that $$ \int_n^{n+1}\frac{\mathrm{d}x}{\sqrt{x}}\le\frac1{\sqrt{n}}\le\int_{n-1}^n\frac{\mathrm{d}x}{\sqrt{x}} $$ Adding up the integrals is simple.

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Approach $\bf{2}$

Another way is to note that $$ 2\sqrt{n+1}-2\sqrt{n\vphantom{+1}}=\frac2{\sqrt{n+1}+\sqrt{n\vphantom{+1}}} $$ That is, $$ 2\sqrt{n+1}-2\sqrt{n\vphantom{+1}}\le\frac1{\sqrt{n\vphantom{+1}}}\le2\sqrt{n\vphantom{-1}}-2\sqrt{n-1} $$ The bounds can be summed as a telescoping series.

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Approach $\bf{3}$ $$ \begin{align} \frac1{\sqrt{n}}-2\sqrt{n\vphantom{-1}}+2\sqrt{n-1} &=\frac1{\sqrt{n}}-\frac2{\sqrt{n\vphantom{-1}}+\sqrt{n-1}}\\ &=\frac{\sqrt{n-1}-\sqrt{n\vphantom{-1}}}{n+\sqrt{n(n-1)}}\\[6pt] &\lt0 \end{align} $$ This implies that the sequence $$ a_n=\sum_{k=1}^n\frac1{\sqrt{k}}-2\sqrt{n} $$ is decreasing. $$ \begin{align} \frac1{\sqrt{n}}-2\sqrt{n+1}+2\sqrt{n\vphantom{+1}} &=\frac1{\sqrt{n}}-\frac2{\sqrt{n+1}+\sqrt{n\vphantom{+1}}}\\ &=\frac{\sqrt{n+1}-\sqrt{n\vphantom{+1}}}{n+\sqrt{n(n+1)}}\\[6pt] &\gt0 \end{align} $$ This implies that the sequence $$ b_n=\sum_{k=1}^n\frac1{\sqrt{k}}-2\sqrt{n+1} $$ is increasing.

Note that $$ a_n-b_n=\frac2{\sqrt{n\vphantom{+1}}+\sqrt{n+1}} $$ which tends to $0$.

Thus any $b_n$ is a lower bound for $a_n$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With a Riemann Zeta Function Identity:

\begin{align} \mrm{f}\pars{n} & \equiv 1 + 1/\root{2} + 1/\root{3} + \cdots + 1/\root{n} - 2\root{n} = \sum_{k = 1}^{n}{1 \over k^{\color{red}{1/2}}} - 2\root{n} \\[5mm] & = \bracks{\zeta\pars{\color{red}{1 \over 2}} - {n^{1 - \color{red}{1/2}} \over \color{red}{1/2} - 1} + \color{red}{1 \over 2}\int_{n}^{\infty}{x - \left\lfloor\,{x}\,\right\rfloor \over x^{\color{red}{1/2} + 1}}\,\dd x} - 2\root{n} \\[5mm] & = \zeta\pars{1 \over 2} + {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x \end{align}

Note that $\ds{0 < {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x < {1 \over 2}\int_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\LARGE\to}\,\,\,\color{red}{\large 0}}$

$$ \bbx{\mbox{Then,}\quad\lim_{n \to \infty}\mrm{f}\pars{n} = \zeta\pars{1 \over 2}} \approx -1.4604 $$