I need to calculate limit:

$$ \lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{2n}}- \frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+2}}-\dotsb+\frac{1}{\sqrt{4n}}\right ) $$

Any hints how to do that would be appreciated.

  • $\begingroup$ Your summation is unclear to me. Could you either add more terms or explicitly state the summands in a formula? $\endgroup$ – Zach466920 Mar 25 '15 at 18:47
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    $\begingroup$ Are you sure that the last value of the sum has negative sign? $\endgroup$ – Prasun Biswas Mar 25 '15 at 18:51
  • $\begingroup$ @ Prasun Biswas, Yes. $\endgroup$ – tina tulip Mar 25 '15 at 19:07
  • $\begingroup$ @tinatulip, but according to the first few terms, all the terms with even value inside square root have positive sign and the ones with odd value inside square root have negative sign. $4n$ is even, so that term should have negative sign to follow the general term $a_n=(-1)^n \dfrac{1}{\sqrt{n}}$. The given sum inside limit is the sum of the elements of this sequence: $\{a_i\}_{i=2n}^{i=4n}$ $\endgroup$ – Prasun Biswas Mar 25 '15 at 19:37
  • $\begingroup$ For example: $$n = 1: \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}}$$ $$n=2: \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{6}} - \frac{1}{\sqrt{7}} + \frac{1}{\sqrt{8}}$$ $\endgroup$ – Joel Mar 25 '15 at 19:46

The quantity inside the limit is between $\frac{1}{\sqrt{2n}}$ and $\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}$, hence the limit is zero by squeezing. To notice it, it is sufficient to couple consecutive terms: let $A_n=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$.

Then $A_n>0$ and: $$ \frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\ldots+\frac{1}{\sqrt{4n}}=\frac{1}{\sqrt{2n}}-\sum_{k=n}^{2n-1}A_{2k+1}<\frac{1}{\sqrt{2n}} $$ as well as: $$ \frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\ldots+\frac{1}{\sqrt{4n}}=A_{2n}+\sum_{k=n+1}^{2n-1}A_{2k}+\frac{1}{\sqrt{4n}}>\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}.$$

  • $\begingroup$ @Jack D'Aurizio, thank you. But how can i conclude that the limit is between $\frac{1}{\sqrt{2n}}$ and $\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}$ ? $\endgroup$ – tina tulip Mar 25 '15 at 19:00
  • $\begingroup$ @tinatulip: $\left\{\frac{1}{\sqrt{n}}\right\}$ is a decreasing sequence, hence it is enough to "couple" consecutive terms. $\endgroup$ – Jack D'Aurizio Mar 25 '15 at 20:49
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    $\begingroup$ Why the quantity is inside these two terms. This is extremely hard to see. $\endgroup$ – user139708 Mar 26 '15 at 2:43
  • $\begingroup$ @ProbabilityGuy: or sum can be written as $$\frac{1}{\sqrt{2n}}-\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+2}}\right)-...-\left(\frac{1}{\sqrt{4n-1}}-\frac{1}{\sqrt{4n}}\right)$$ $\endgroup$ – Jack D'Aurizio Mar 26 '15 at 12:52
  • $\begingroup$ or as $$\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}+\left(\frac{1}{\sqrt{2n+2}}-\frac{1}{\sqrt{2n+3}}\right)+\ldots -\frac{1}{\sqrt{4n}}$$ Is it clear now? $\endgroup$ – Jack D'Aurizio Mar 26 '15 at 12:54

The limit you are asking for appears to be:

$$\lim_{n\to\infty} \sum_{m=2n}^{4n} \frac{(-1)^m}{\sqrt{m}}.$$

Note that the series $$\sum_{n=0}^\infty \frac{(-1)^m}{\sqrt{m}}$$ converges by the alternating series test. A series converges if for every $\epsilon > 0$ there exists an integer $N$ for which given any $n_2>n_1 > N$ we have

$$\left| \sum_{m=n_1}^{n_2} \frac{(-1)^m}{\sqrt{m}} \right| < \epsilon.$$

We can choose $n_1 = 2n$ and $n_2=4n$. Thus for any $\epsilon>0$, $$\left| \sum_{m=2n}^{4n} \frac{(-1)^m}{\sqrt{m}} \right| < \epsilon$$ for sufficiently large $n$. Since $\epsilon$ can be as small as we want, the limit must be zero.

  • $\begingroup$ does the series converge to $0$? How can you decude it? I really didnt understand. $\endgroup$ – vudu vucu Mar 25 '15 at 19:00
  • $\begingroup$ @vuduvucu The series doesn't converge to zero. The limit that the OP asked for was actually just a sequence of the remainder terms for a series that converges. Necessarily the remainders must tend to zero by the cauchy criterion. $\endgroup$ – Joel Mar 25 '15 at 19:05

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