Linked Questions

35 votes
12 answers
67k views

Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? [duplicate]

Can you please explain why $$ \sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2 $$ I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{...
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  • 393
7 votes
10 answers
2k views

Why does $\sum_{n = 0}^\infty \frac{n}{2^n}$ converge to 2? [duplicate]

Apparently, $$ \sum_{n = 0}^\infty \frac{n}{2^n} $$ converges to 2. I'm trying to figure out why. I've tried viewing it as a geometric series, but it's not quite a geometric series since the ...
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12 votes
2 answers
128k views

Sum of a power series $n x^n$ [duplicate]

I would like to know: How come that $$\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}$$ Why isn't it infinity?
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18 votes
4 answers
37k views

How to calculate: $\sum_{n=1}^{\infty} n a^n$ [duplicate]

I've tried to calculate this sum: $$\sum_{n=1}^{\infty} n a^n$$ The point of this is to try to work out the "mean" term in an exponentially decaying average. I've done the following: $$\text{let }...
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  • 473
8 votes
6 answers
1k views

$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum [duplicate]

Possible Duplicate: How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$ How can you compute the limit of $\sum \limits_{n=1}^{\infty} n(2/3)^n$ Evidently it is equal to 6 by wolfram alpha ...
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  • 2,023
6 votes
6 answers
2k views

Sequence sum question: $\sum_{n=0}^{\infty}nk^n$ [duplicate]

I am very confused about how to compute $$\sum_{n=0}^{\infty}nk^n.$$ Can anybody help me?
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4 votes
5 answers
14k views

Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ [duplicate]

I have tried to compute the first few terms to try to find a pattern but I got $$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$ but I still don't see any obvious ...
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  • 119
3 votes
6 answers
248 views

Sum of infinite series $1+\frac22+\frac3{2^2}+\frac4{2^3}+\cdots$ [duplicate]

How do I find the sum of $\displaystyle 1+{2\over2} + {3\over2^2} + {4\over2^3} +\cdots$ I know the sum is $\sum_{n=0}^\infty (\frac{n+1}{2^n})$ and the common ratio is $(n+2)\over2(n+1)$ but i ...
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4 votes
5 answers
10k views

Formula for $r+2r^2+3r^3+...+nr^n$ [duplicate]

Is there a formula to get $r+2r^2+3r^3+\dots+nr^n$ provided that $|r|<1$? This seems like the geometric "sum" $r+r^2+\dots+r^n$ so I guess that we have to use some kind of trick to get it, but I ...
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  • 2,817
5 votes
3 answers
522 views

Evaluate the series $\sum _{n=1}^{\infty} \frac{n}{5^n}$ [duplicate]

$$\sum _{n=1}^{\infty}\frac{n}{5^n}$$ I tried to plug in $n=1,2,3,4,...$ but I can't use common ratio to solve problem. I think there is another way like using differentiation or integral but I don't ...
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  • 53
2 votes
3 answers
2k views

Proving $ \sum_{n=1}^{\infty} nz^{n} = \frac{z}{(1-z)^2}$ for $z \in (-1, 1)$ [duplicate]

I do not know where to start, any hints are welcome.
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  • 2,429
0 votes
5 answers
7k views

What is the sum of the series 1/3 + 2/9 + 3/27 + 4/81 + ........ [duplicate]

I remember solving this in highschool , but now I don't remember how to find sum of these kind of series . I want to find the sum of the general series Sum $\sum_{n=1}^{\infty} n .a^{-n} = ? $ ...
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2 votes
4 answers
260 views

Question on a tricky Arithmo-Geometric Progression:: [duplicate]

$$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty$$ This summation was irritating me from the start,I don't know how to attempt this ,tried unsuccessful attempts ...
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  • 517
2 votes
2 answers
7k views

Find value of infinite sum [duplicate]

Possible Duplicate: How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$ How would I go about deriving the value of the following infinite sum: $\sum\limits_{k=1}^\infty kx^k$ ? I thought ...
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  • 117
4 votes
3 answers
421 views

Sum of Arithmetic-Geometric Series [duplicate]

So I am having trouble getting the sum of the series: $1 + 2\left(\frac32\right) + 3\left(\frac{3^2}{2^2}\right) + ... + k\left(\frac{3^{k-1}}{2^{k-1}}\right)$ I cant figure out for the life of me ...
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