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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.

José Carlos Santos
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user639336
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  • connected : https://math.stackexchange.com/q/1432335 – Jean Marie May 02 '19 at 16:29
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    Just for giggles, the limit is $\zeta(1/2)$. – robjohn May 02 '19 at 21:36
  • @robjohn Not "just for giggles" : it is a very valuable remark because it connects this issue with a fundamental function which is behind the curtain, here in the framework of its analytical continuation. – Jean Marie 8 mins ago – Jean Marie May 03 '19 at 06:50
  • @JeanMarie: I said "just for giggles", since the methods I know to evaluate this limit are most likely beyond the scope of this question. – robjohn May 03 '19 at 07:32

5 Answers5

4

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.)

Sangchul Lee
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  • @Lee sir I really cant understand, why your computaion follow that f(n) converges? – user639336 May 02 '19 at 15:26
  • @Lee sir please help me – user639336 May 02 '19 at 15:30
  • @user639336, Before answering your question, have you ever heard of $p$-series test, which tells that $$\sum_{n=1}^{\infty} \frac{1}{n^p} $$ converges if and only if $ p > 1$? The detail of the answer will depend on whether you know this or not. – Sangchul Lee May 02 '19 at 15:37
  • @Lee sir absolutely I'm aware about it, but it forma p series cant get it – user639336 May 02 '19 at 15:39
  • @Lee sir I'm looking for that if it is p series, but cant find how it is p series? – user639336 May 02 '19 at 15:40
  • @user639336, It is not exactly a $p$-series, but the catch is that you can give a bound which comes from a $p$-series. Indeed, using $\sqrt{k}+\sqrt{k-1} \geq \sqrt{k}$, we get $$f(n) \geq -\sum_{k=1}^{n} \frac{1}{k^{3/2}} \geq -\sum_{k=1}^{\infty} \frac{1}{k^{3/2}}. $$ – Sangchul Lee May 02 '19 at 15:42
  • @Lee sir thank you so much. If you kindly help me how it (as a sequence) is bdd below then it would be great, because my approach was to prove m.d & bdd below, the 1st one is done but the problem is the 2nd one as I mentioned in my question. – user639336 May 02 '19 at 15:57
  • @user639336 I'm not even sure if it's decreasing, but it doesn't matter. If you use the $p-$series test, you're done. – Ovi May 02 '19 at 16:15
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    $$g(n)=1+1/\sqrt2+1/\sqrt3+\cdots +1/\sqrt n-2\sqrt { n +1 }$$ is smaller and increasing, the two come together at a common limit – Will Jagy May 02 '19 at 19:09
  • Sangchul, I put in a table. I think the two sequence argument, where they must come together, has an emotional appeal for beginners. Meanwhile, years ago I saw a quote on the wall at UCLA, probably by Francis Bacon, roughly comparing mathematics and science. Is it still there? – Will Jagy May 02 '19 at 19:20
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    maybe this: It has come to pass, I know not how, that Mathematics and Logic, which ought to be but the handmaids of Physic, nevertheless presume on the strength of the certainty which they possess to exercise dominion over it. – Will Jagy May 02 '19 at 19:26
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    @Will Jagy Hi Will. I appreciate this philosophy touch... somewhat too scarce on this site. – Jean Marie May 03 '19 at 06:42
  • @Sangchul Lee You mention Euler MacLaurin expansion. I have attempted to do such an expansion for $1/\sqrt{x}$ on $[0,1]$ but it has been unsuccessful (I have deleted my answer) due to the fact that this function is unbounded at bound $0$... Your comment ? – Jean Marie May 03 '19 at 06:48
  • @WillJagy, That is a very nice idea! Anyway, I even did not know there are writings on the wall... You certainly know better about this university than me :) – Sangchul Lee May 03 '19 at 15:49
  • @JeanMarie, The idea of Euler-MacLaurin expansion is to approximate the sum $\sum_{k=1}^{n} f(k)$ by $\int_{1}^{n} f(x) , \mathrm{d}x$, so you may be more interested in the interval $[1, n]$ than $[0, 1]$. – Sangchul Lee May 03 '19 at 15:51
  • @Sangchul Lee Thank you very much. You are right. I realize that I was using an interval adapted to a Riemann/Darboux sum, not to a Euler-Maclaurin expansion. – Jean Marie May 03 '19 at 16:30
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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)
Will Jagy
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  • sir I want to prove it is bounded below and also want to see its bounds, you are showing by table, I want concrete proof. – user639336 May 03 '19 at 02:12
  • @user639336 and I want you to prove one sequence is decreasing and the other is increasing. – Will Jagy May 03 '19 at 02:33
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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 $\mathbb R$.

Piquito
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  • sir why there 2*sqrt(x) survive in your first term? Is'nt that term will vanish with the before? – user639336 May 02 '19 at 17:01
  • Write separately $f(x+h)$ and $f(x)$ then $f(x+h)-f(x)$. You will clearly see why. – Piquito May 02 '19 at 20:28
  • sir yes and thanks, but sir why it is enough to show lim(x->∞){2sqrt(x)-2sqrt(x+h)} =0? sir from your f(x+h)-f(x) I want to show |f(x+h)-f(x)|<epsilon but I'm unable to do so, please sir help me here. – user639336 May 03 '19 at 03:56
  • You can make, for example $\sqrt{x}-\sqrt{x+h}=\sqrt x(1-\sqrt{1+\frac hx})$. A way is that $1-\sqrt{1+\frac hx}$ tends faster to $0$ than $\sqrt x$ tends to $\infty$. However it is rather convenient for you maybe to put $$\sqrt{x}-\sqrt{x+h}=\frac{1-\sqrt{1+\dfrac hx}}{\dfrac{1}{\sqrt x}}$$ and apply L'Hôpital' s rule. – Piquito May 04 '19 at 14:13
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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.

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.

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 $a_n$.

robjohn
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  • sir how we can get bounds by telescoping in your approach 2? – user639336 May 03 '19 at 02:37
  • $$ \sum_{k=1}^n\left(2\sqrt{k+1}-2\sqrt{k\vphantom{+1}}\right)\le\sum_{k=1}^n\frac1{\sqrt{k}}\le\sum_{k=1}^n\left(2\sqrt{k\vphantom{-1}}-2\sqrt{k-1}\right) $$ Telescoping Sums yields $$ 2\sqrt{n+1}-2\le\sum_{k=1}^n\frac1{\sqrt{k}}\le2\sqrt{n} $$ Therefore, $$ -2\le\sum_{k=1}^n\frac1{\sqrt{k}}-2\sqrt{n}\le0 $$ – robjohn May 03 '19 at 03:04
  • Note that $\zeta!\left(\frac12\right)\doteq-1.4603545088$ – robjohn May 03 '19 at 03:06
  • [+1] Very thorough answer ! Besides, how can it be established that this limit is indeed $\zeta(\tfrac12)$ ? I have tried using $\zeta(\tfrac12)=(1+\sqrt{2})\sum_{k=1}^{\infty} \dfrac{(-1)^k}{\sqrt{k}}$ without success... – Jean Marie May 03 '19 at 06:32
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    @JeanMarie: this, and more, can be shown using the Euler-Maclaurin Sum Formula. For $\mathrm{Re}(z)\gt0$, the limit $$\zeta(z)=\lim\limits_{n\to\infty}\left[\sum_{k=1}^n\frac{1}{k^z}-\frac{1}{1-z}n^{1-z}\right]$$ is shown in this answer. Plug in $z=\frac12$. – robjohn May 03 '19 at 07:25
  • @robjohn Thank you very much. – Jean Marie May 03 '19 at 07:41
  • @JeanMarie: Suppose we know that $$\lim_{n\to\infty}\left[\sum_{k=1}^n\frac1{\sqrt{k}}-2\sqrt{n}\right]=\alpha$$ Then $$\lim_{n\to\infty}\left[\sum_{k=1}^{2n}\frac1{\sqrt{k}}-2\sqrt{2n}\right]=\alpha$$ and $$\lim_{n\to\infty}\left[\sum_{k=1}^{n}\frac1{\sqrt{2k}}-\frac{2\sqrt{n}}{\sqrt{2}}\right]=\frac\alpha{\sqrt{2}}$$ Subtract twice the third from the second and take the limit $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{\sqrt{k}}=\left(1-\sqrt{2}\right)\alpha$$ which is the same as $$\left(1+\sqrt{2}\right)\sum_{k=1}^\infty\frac{(-1)^k}{\sqrt{k}}=\alpha$$ – robjohn May 03 '19 at 09:15
  • @robjohn Thank you very much : a pretty computation... – Jean Marie May 03 '19 at 09:19
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

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

Felix Marin
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