Point of convergence of series

Question

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How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$

How do I compute the point of convergence of a series?

e.g. for example, How do I prove that,

$$\sum_{n=0}^{\infty}\frac{n}{2^{n+1}}=1$$

Can I do:

Point of convergence= $\limsup_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{n}{2^{n+1}}=\sup\{\frac{1}{4},\frac{1}{2},\frac{11}{16},\frac{13}{16},\frac{57}{64},\frac{15}{16},\dots\}=1$

Here is a (near) duplicate post. – David Mitra Jan 15 '13 at 19:09 — David Mitra, Jan 15 '13 at 19:09
Thanks! That was really helpful! – 007resu Jan 15 '13 at 19:17 — 007resu, Jan 15 '13 at 19:17

lab bhattacharjee · Accepted Answer · 2013-01-15T19:29:56.650

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$$\sum_{n=0}^\infty r^n=\frac1{1-r}$$ for $|r|<1$ and $r\ne1$

Applying derivative wrt $r,$ $$\sum_{n=0}^\infty nr^{n-1}=\frac1{(1-r)^2}$$

Put $r=\frac12,$ $$\sum_{n=0}^\infty \frac{n}{2^{n-1}}=\frac1{(1-\frac12)^2}=4$$

Now, divide both sides by $2^2$

Alternatively,

let $$S=1+2r+3r^2+4r^3+\cdots+ n\cdot r^{n-1}$$ with $|r|<1$ and $r\ne1$

So, $$r\cdot S=r+2r^2+3r^3+4r^4+\cdots+(n-1)\cdot r^{n-1}+n\cdot r^n$$

On subtraction, $$(1-r)S=1+r+r^2+r^3+\cdots+r^{n-1}-n\cdot r^n=\frac{1-r^n}{1-r}-n\cdot r^n$$

If $n\to \infty, r\to 0$ as $|r|<1$ , so does $n\cdot r^n$(Prove)

So, $$(1-r)S=\frac1{1-r}$$

edited Jan 15 '13 at 19:29

answered Jan 15 '13 at 18:59

lab bhattacharjee

274,582

Wonderful! Neat answer, lab. +1 – Mikasa Jan 15 '13 at 19:04
@BabakSorouh, my pleasure. Please have a look into the alternative method. – lab bhattacharjee Jan 15 '13 at 19:07
Well, the second method is risky right! you never know when 1=2=1/2... – 007resu Jan 15 '13 at 19:08
How do you draw that line just below "divide both sides by $2^2$"? – Git Gud Jan 15 '13 at 19:08
@GitGud, Do you know, you can click edit to see the actual codes? "*****" is what he did. – 007resu Jan 15 '13 at 19:09
@GitGud, as each term after derivative is $n\cdot r^{n-1}$ but we need $n\cdot r^{n+1}$ – lab bhattacharjee Jan 15 '13 at 19:09
@Freddy, what did you mean by $1=2=\frac12?$ – lab bhattacharjee Jan 15 '13 at 19:10
@Freddy Thanks. @ labbhattacharjee, You misunderstood my question. I got it now anyway. Thanks. – Git Gud Jan 15 '13 at 19:12
@labbhattacharjee, I realized, its fine here. What I was refering to 1-1+1-1+1... and the old trick of showing that 0=1=1/2 or something. But we have convergent series and it's fine. – 007resu Jan 15 '13 at 19:17
@DavidMitra, thanks for your instructive feedback, rectified. – lab bhattacharjee Jan 15 '13 at 19:40

user 1591719 · Answer 2 · 2013-01-15T19:26:34.647

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If we denote S=$\sum_{n=0}^{\infty}\displaystyle \frac{n}{2^{n+1}}$, then

$$S-\frac{S}{2}=\sum_{n=2}^{\infty}\frac{1}{2^k}\longrightarrow S=1$$

Q.E.D.

edited Jan 15 '13 at 19:26

answered Jan 15 '13 at 19:21

user 1591719

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It's really clever method! Thanks! – 007resu Jan 15 '13 at 19:29
@Freddy: thank you! :-) – user 1591719 Jan 15 '13 at 19:29

Point of convergence of series

2 Answers2