Finding the sum of a series

[steven187]

hello all

you know we have all these tests for convergence of a series, but it always made me wonder if there exists any other method of finding the sum of a series, like we have the geometric series in which we are able to find the sum by a simple formula, but are we able to find such a formula for any series?, would such a formula be unique to each particular series?, or to a group of such series?

for example $$\sum_{n=1}^{\infty}\frac {(-2)^{n}}{n+1}$$
this isn't a geometric series then how would you find the sum of such series?
is it possible to derive a formula for any series that exist, or are there limitations or conditions that need to be satisfied?

 

[shmoe]

There is no sure fire general way to find the sum of an infinite series in a 'nice' form. If a series converges and you have enough time on your hands you can find as many decimal places as you like, but an exact, nice looking solution is not always possible (indeed your series might not even converge to something 'nice').

The series you've provided doesn't converge by the way, the terms don't even go to 0 (a necessary but not sufficient condition).

 

[steven187]

so your sayin that it is possible in some cases to derive a formula for a series besides a geometric series but in most cases its not possible? , and about that series it was suppose to be

$$\sum_{n=1}^{\infty}\frac {(-2)^{-n}}{n+1}$$
like would i be able to derive a formula for this series? or how would i be able to find the sum of such a series?

 

[saltydog]

It's a very interesting question. There are various techniques for solving such. For example,

$$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$

This can be solved by using Fourier coefficients and Parseval's Theorem.

$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}k=ln(2)$$

is shown by considering:

$$f(x)=ln(1+x)$$

and differentiating and considering the Taylor series (thanks Daniel).

$$\sum_{k=1}^{\infty}\frac{1}{4k(2k-1)}$$

is solved by considering the sum:

$$S=\frac{1}{2}[(1-1/2)+(1/3-1/4)+(1/5-1/6)+...$$

and:

$$\sum_{n=0}^{\infty} ne^{-an}$$

is solved by considering:

$$z=\sum_{n=0}^{\infty}w^n$$

with:
$$w=e^{-a}$$

and differentiating both the sum and the expression for the sum of the corresponding geometric series with respect to w.

Tons more I bet. Would be nice to have a compilation of the various methods for calculating infinite sums.

 

[shmoe]

so your sayin that it is possible in some cases to derive a formula for a series besides a geometric series but in most cases its not possible?

Yes. saltydog's given numerous examples. Telescoping series is another handy one, so is recognizing your sum as a known power series, rearranging terms sometimes helps, and more .

, and about that series it was suppose to be

$$\sum_{n=1}^{\infty}\frac {(-2)^{-n}}{n+1}$$
like would i be able to derive a formula for this series? or how would i be able to find the sum of such a series?

Relate it to the power series for log(1+x).

 

[HallsofIvy]

so your sayin that it is possible in some cases to derive a formula for a series besides a geometric series but in most cases its not possible? , and about that series it was suppose to be

$$\sum_{n=1}^{\infty}\frac {(-2)^{-n}}{n+1}$$
like would i be able to derive a formula for this series? or how would i be able to find the sum of such a series?

Some times you can recognize a series as a special case of a Taylor's series for a function. To sum the particular series you give here, I would recall that the Taylor's series for ln(x+1) is $$\sum_{n=1}^{\infty}(-1)^n \frac{x^n}{n}$$ (and converges for -1< x< 1). That differs from your series on having n+1 in the denominator: Okay, change the numbering slightly. If we let i= n+1 then n= i-1 and your series becomes $$\sum_{i=2}^{\infty}\frac{(-2)^{-i+1}}{i}$$. I can just take that (-2)¹ out of the entire series: $$(-2)\sum_{i=2}^{\infty}\frac{(-2)^{-i}}{i}$$. We're almost there! Since I need (-2)^-i instead of xⁱ, take x= -1/2 which is in the radius of convergence. The fact that the sum now starts at i= 2 instead of 1 is not a problem: just calculate that separately- when i= 0 the term is (-2)⁰/1= -2: Your sum is $$(-2)\sum{i=1}\frac{(-1)^i}{i}+ 2)= (-2)ln(-1/2+ 1)+ 2= -2ln(1/2)+ 2= 2+ 2 ln(2)= 2(1+ ln(2))$$.

 

[saltydog]

Using Hall's method I obtained a different answer:

For this sum:

$$\sum_{n=1}^{\infty}\frac{(-2)^{-n}}{n+1}$$

Consider:

$$ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}\quad\text{for}\quad-1<x<1$$

Letting $x=\frac{1}{2}$

$$ln(3/2)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^n n}$$

Letting i=n-1 we obtain:

$$\begin{align*} ln(3/2)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^n n}&=\sum_{i=0}^{\infty}\frac{(-1)^{i+2}}{2^{i+1} (i+1)}\\ &=\frac{1}{2}+\sum_{i=1}^{\infty}\frac{(-1)^2 (-1)^i}{(2)2^i(i+1)}\\ &=\frac{1}{2}+\frac{1}{2}\sum_{i=1}^{\infty}\frac{(-1)^i}{2^i(i+1)}\\ &=\frac{1}{2}+\frac{1}{2}\sum_{i=1}^{\infty}\frac{(-2)^{-i}}{i+1} \end{align}$$

Solving for the series, I obtain:

$$\sum_{i=1}^{\infty}\frac{(-2)^{-i}}{(i+1)}=-1+2ln(3)-2ln(2)$$

This result agrees with Mathematica.

 

[shmoe]

Solving for the series, I obtain:

$$\sum_{i=1}^{\infty}\frac{(-2)^{-i}}{(i+1)}=-1+2ln(3)-2ln(2)$$

This result agrees with Mathematica.

This is correct, at least it's also what I had. Halls has a couple small mistakes, he has the series for -log(1+x), not log(1+x), and used x=-1/2 when it should have been x=1/2. Also the 'missing term' is then $$(-2)(1/2)^1/1=-1$$.

Notice it's an alternating series with decreasing terms and the first term is negative, therefore the sum must also be negative. You could go furthur if you like, the sum must be larger than $$-\frac{1}{2\times 2}=-\frac{1}{4}$$ and smaller than $$-\frac{1}{2\times 2}+\frac{1}{2^2\times 3}=-\frac{1}{6}$$, and so on if you wanted some more reassurance that your answer was correct (an alternating series with decreasing terms will hop back and forth between being greater and less than the sum). Of course this sort of estimate won't prove your answer is correct, but can catch mistakes.

 

[steven187]

hello all

now I realize that methods of finding the sum of an infinite series is limited,
so all we have is, telescoping ,geometric series, rearranging terms (could someone give me an example of this thanxs), manipulation, Fourier series- (which i can bearly remember) is there anymore?

also what is the name of the method saltydog was originally using? I didnt really understand it but i have to admit anything unusual is interesting, saltydog please help? thanxs

well after making up a few unusual series and trying to derive the sum, for the majority of them i couldn't derive anything except for one of them , and here it is please update me on what you think?

$$\sum_{k=0}^{\infty}\left(\begin{array}{cc}k \\r \end{array}\right)x^{k-r}=\frac{1}{(1-x)^{r+1}}$$ took me ages to get this result, I shall call it the steven zeta series lol.
does anybody have any other examples of such results?

 

[saltydog]

saltydog please help? thanxs

Steven . . . I struggle with these and a lot of other things in the forum. I made sure to give Hall credit above for that method. He suggested the technique and I followed through. I take 1 point credit for that work. Really I suspect there is some compilation which goes into more detail with sums. I would expect that even infinite sums could/should/would be arranged into groupings similar to integrals and differential equations and solved using specific technique applicable to membership in a particular class like we do for integrals. Anyway, I think if someone intensively investigated them this could be done to some measure of success. Interesting problems you bring up here though.

 

[steven187]

hello all

well i didnt really understaand what you ment by this, what is the problem being solved here, or what method is it?

.

$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}k=ln(2)$$

is shown by considering:

$$f(x)=ln(1+x)$$

and differentiating and considering the Taylor series (thanks Daniel).

$$\sum_{k=1}^{\infty}\frac{1}{4k(2k-1)}$$

is solved by considering the sum:

$$S=\frac{1}{2}[(1-1/2)+(1/3-1/4)+(1/5-1/6)+...$$

and:

$$\sum_{n=0}^{\infty} ne^{-an}$$

is solved by considering:

$$z=\sum_{n=0}^{\infty}w^n$$

with:
$$w=e^{-a}$$

and differentiating both the sum and the expression for the sum of the corresponding geometric series with respect to w.

well now Iv been searching the net for ages to find some kind of summary of all the different types of series, methods of finding the sum of a series, and special cases of series with there sums like the power, euler, Laurent, alternating ,arithmetic, hypergeometric series, Maclaurin ,binomial, taylor, harmonic, riemann, geometric, fourier, pathological,Asymptotic Series, Dirichlet L-series, Multiple Series, Hyperasymptotic Series and Superasymptotic Series( i don't think there is anymore) oh yes and the steven zeta series, well out of hours of searching i only found one document but that wasnt even good enough, does anybody have any links?, well saltydog I hope that there is some kind of compilation or else it looks like I might have to compile my own,
well as to the original problem this is how i did it enjoy

$$\int_0^x\sum_{n=0}^{\infty}z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\int_0^x z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log{(1-x)}$$
now we substitute $$x=\frac{-1}{2}$$ in which after manipulating it we get
$$\sum_{n=1}^{\infty}\frac{(-2)^{-n}}{(n+1)}=-1+2\log[\frac{3}{2}]$$

 

[TenaliRaman]

$$\int_0^x\sum_{n=0}^{\infty}z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\int_0^x z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log{(1-x)}$$
now we substitute $$x=\frac{-1}{2}$$ in which after manipulating it we get
$$\sum_{n=1}^{\infty}\frac{(-2)^{-n}}{(n+1)}=-1+2\log[\frac{3}{2}]$$

I believe you realize that your first few statements are to be justified even though they are correct.

-- AI

 

[saltydog]

$$\int_0^x\sum_{n=0}^{\infty}z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\int_0^x z^n dz=\int_0^x \frac{1}{1-z} dz$$
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log{(1-x)}$$
now we substitute $$x=\frac{-1}{2}$$ in which after manipulating it we get
$$\sum_{n=1}^{\infty}\frac{(-2)^{-n}}{(n+1)}=-1+2\log[\frac{3}{2}]$$

Thanks Steven. That's very nice. Tenali, you mind explaining what needs to be justified?

 

[steven187]

hello

thanxs saltydog, well about what needs to be justified is that to swap the summation sign and the integral sign it needs to satisfy the conditions of the dominated convergence theorem i don't think there is anythin else to justify

Steven