Can I use limits to calculate summation of a progression

Question

I had a question

$$\sum_{a=1}^n \frac{1}{a^2}=?$$

I had learned newton's method of undetermined method ,but that doesn't work here because of negative power

than I saw another question

$$\sum_{a=1}^n \frac{1}{a(a+1)}$$

here we transformed it as

$$\sum_{a=1}^n (\frac{1}{a}-\frac{1}{a+1})$$

which on expanding gave

$$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{n}-\frac{1}{n+1}$$

where each element cancels out except

$$1-\frac{1}{n+1}$$

which gives the formula

$$\sum_{a=1}^n \frac{1}{n(n+1)}=1-\frac{1}{n+1}=\frac{n}{n+1}$$

but I can't express $a^{-2}$ as two diffrent factors so I thought to use limits(calculus) as

$$\frac{1}{a^2}=\lim_{h\to 0}\frac{1}{a(a+h)}$$

but when I tried to break it then I got $h$ in denominator.

I may be wrong, so please help to calculate the summation.

You may take a look here https://en.wikipedia.org/wiki/Harmonic_number#Generalized_harmonic_numbers — Marco Cantarini, Dec 24 '16 at 15:08
HERE is another way the evaluate the series that relies on real analysis only. — Mark Viola, Dec 24 '16 at 16:13

Simply Beautiful Art · Answer 1 · 2016-12-25T15:56:11.643

4

There isn't such a beautiful solution as the one you've presented, since it is the case that as $n\to\infty$, you get the Basel problem.

Notice that

$$\frac1{a(a+h)}=\frac1h\left(\frac1a-\frac1{a+h}\right)$$

but since $h$ is not a whole number, we will not be seeing any cancellations like with your telescoping series.

But, if you could recall the geometric series:

$$\frac{1-r^{n+1}}{1-r}=\sum_{k=0}^nr^k$$

Integrate both sides and divide by $x:$

$$\frac1x\int_0^x\frac{1-r^{n+1}}{1-r}\ dr=\frac1x\int_0^x\sum_{k=0}^nr^k\ dr=\sum_{k=0}^n\frac1{k+1}x^k$$

Repeating the process,

$$\int_0^y\frac1x\int_0^x\frac{1-r^{n+1}}{1-r}\ dr\ dx=\int_0^y\sum_{k=0}^n\frac1{k+1}x^k\ dx=\sum_{k=0}^n\frac1{(k+1)^2}y^{k+1}$$

And as $y\to1$,

$$\sum_{k=1}^nk^{-2}=\int_0^1\frac1x\int_0^x\frac{1-r^n}{1-r}\ dr\ dx$$

I believe I've read that this is how Euler approximated the Basel problem.

edited Dec 25 '16 at 15:56

answered Dec 24 '16 at 15:06

Simply Beautiful Art

74,685

You are genius. :) :) :) :). Not sarcasm , I m serious. +1 – Vidyanshu Mishra Dec 24 '16 at 15:12
@THELONEWOLF. :( but wait, I wasn't even done yet! – Simply Beautiful Art Dec 24 '16 at 15:13
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That is why said that. :) :) – Vidyanshu Mishra Dec 24 '16 at 15:13
@Simple Art: i am a beginer in calculus so i would like if you would also provide a easy method to me. – Blaise Thunderstorm Dec 24 '16 at 15:16
but answers by using calculas are also useful to me to learn it in a better way – Blaise Thunderstorm Dec 24 '16 at 15:19
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@BlaiseThunderstorm I cannot imagine a simpler way. Indeed, if there were a simpler way, it wouldn't have taken like 100 years for someone to solve the problem $$\sum_{k=1}^\infty k^{-2}=\frac{\pi^2}6$$ – Simply Beautiful Art Dec 24 '16 at 15:19
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@SimpleArt https://arxiv.org/pdf/1502.07667v1.pdf =) – MathematicsStudent1122 Dec 24 '16 at 15:27
@MathematicsStudent1122 It does not give a very explicit form for $\xi$ though. It simply says that it exists and that it is seemingly unimportant for the limit. – Simply Beautiful Art Dec 24 '16 at 15:29
HERE is another way the evaluate the series that relies on real analysis only. – Mark Viola Dec 24 '16 at 16:13
@Dr.MV Yes, but the OP was really interested in the finite sum. – Simply Beautiful Art Dec 24 '16 at 16:15
@SimpleArt Indeed. I thought this might be a useful supplement. – Mark Viola Dec 24 '16 at 16:20

Marsan · Answer 2 · 2016-12-24T16:46:24.057

I am not aware of any formulas for the finite partial sum... however we can compute the limit ! I think the easiest way to compute it is through trigonometric sums/estimations. Here is a sketch of the proof :

By using De Moivre's Formula, you can easily show that, for all real $t$ :

\begin{equation*} \sin(2n+1)t = \sum_{k=1}^{n} \binom{2n+1}{2k+1}(-1)^k \sin^{2k+1}(t) \cos^{2(n-k)}(t) \end{equation*}

Then dividing by $\sin^{2n+1}(t)$, when $t \notin \pi\mathbb{Z}$, you get

\begin{equation*} \frac{\sin(2n+1)t}{\sin^{2n+1}(t)} = P_n({cotan}^{2}(t)). \; \text{where} \: P_n=\sum_{k=1}^{n} \binom{2n+1}{2k+1}(-1)^k X^{n-k} \end{equation*}

For $t \in (0, \frac{\pi}{2})$, the formula above holds and thus $P_n$ vanishes at points $x_k=cotan(t_k)$ where $t_k= \frac{k\pi}{2n+1}$ for $k \in \{1, \cdots, n\}$. Since $P_n$ is of degree $n$, we found all the roots of $P_n$. And by the coefficient/roots relationship, we can assert that :

\begin{equation*} \sum_{k=1}^n x_k = -\frac{-\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{n(2n-1)}{3} \end{equation*}

Then by squaring and inverting the inequalities for $ t \in(0,\frac{\pi}{2})$ : $0<sin(t)\leq t \leq \tan(t)$, we get :

\begin{equation*} {cotan}^2(t) \leq \frac{1}{t^2} \leq 1 + {cotan}^2(t) \end{equation*}

which yields by taking $t=t_k$ and then summing over $k$ :

\begin{equation*} \frac{n(2n-1)}{3} \leq \frac{(2n+1)^2}{\pi^2} \sum_{k=1}^n \frac{1}{k^2} \leq n + \frac{n(2n-1)}{3} \end{equation*}

Thanks to the squeeze Theorem, you may conclude that :

\begin{equation*} \sum_{k=1}^{+ \infty} \frac{1}{k^2} = \frac{\pi^2}{6} \end{equation*}

He's not looking for $\zeta(2)$, he's looking for the finite sum. :-( — Simply Beautiful Art, Dec 24 '16 at 16:40

Can I use limits to calculate summation of a progression

2 Answers2