Linked Questions
264 questions linked to/from How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?
37
votes
12
answers
84k
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)}{...
- 413
19
votes
2
answers
161k
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?
- 391
8
votes
10
answers
3k
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 ...
- 81
20
votes
4
answers
45k
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:
$$\text{let }...
- 503
10
votes
6
answers
2k
views
$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum [duplicate]
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$
Evidently it is equal to 6 by wolfram alpha ...
- 2,091
6
votes
6
answers
3k
views
Sequence sum question: $\sum_{n=0}^{\infty}nk^n$ [duplicate]
I am very confused about how to compute
$$\sum_{n=0}^{\infty}nk^n.$$
Can anybody help me?
- 61
4
votes
5
answers
14k
views
Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ [duplicate]
I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious ...
- 119
3
votes
6
answers
271
views
Sum of infinite series $1+\frac22+\frac3{2^2}+\frac4{2^3}+\cdots$ [duplicate]
How do I find the sum of $\displaystyle 1+{2\over2} + {3\over2^2} + {4\over2^3} +\cdots$
I know the sum is $\sum_{n=0}^\infty (\frac{n+1}{2^n})$ and the common ratio is $(n+2)\over2(n+1)$ but i ...
- 81
4
votes
5
answers
11k
views
Formula for $r+2r^2+3r^3+...+nr^n$ [duplicate]
Is there a formula to get $r+2r^2+3r^3+\dots+nr^n$ provided that $|r|<1$? This seems like the geometric "sum" $r+r^2+\dots+r^n$ so I guess that we have to use some kind of trick to get it, but I ...
- 3,007
5
votes
3
answers
3k
views
Evaluate the series $\sum _{n=1}^{\infty} \frac{n}{5^n}$ [duplicate]
$$\sum _{n=1}^{\infty}\frac{n}{5^n}$$
I tried to plug in $n=1,2,3,4,...$ but I can't use common ratio to solve problem.
I think there is another way like using differentiation or integral but I don't ...
- 53
2
votes
3
answers
3k
views
Proving $ \sum_{n=1}^{\infty} nz^{n} = \frac{z}{(1-z)^2}$ for $z \in (-1, 1)$ [duplicate]
I do not know where to start, any hints are welcome.
- 2,449
0
votes
5
answers
9k
views
What is the sum of the series 1/3 + 2/9 + 3/27 + 4/81 + ........ [duplicate]
I remember solving this in highschool , but now I don't remember how to find sum of these kind of series .
I want to find the sum of the general series
Sum $\sum_{n=1}^{\infty} n .a^{-n} = ? $
...
- 275
2
votes
4
answers
284
views
Question on a tricky Arithmo-Geometric Progression:: [duplicate]
$$\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+\dfrac{5}{64}+\cdots\infty$$
This summation was irritating me from the start,I don't know how to attempt this ,tried unsuccessful attempts ...
- 517
2
votes
2
answers
7k
views
Find value of infinite sum [duplicate]
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
How would I go about deriving the value of the following infinite sum: $\sum\limits_{k=1}^\infty kx^k$ ?
I thought ...
- 117
3
votes
3
answers
465
views
Can you help me solve this series: $\sum_{k=0}^\infty p^kk$? [duplicate]
I have the infinite series:
$$\sum_{k=0}^\infty p^kk$$
where $0<p<1$. I can see from a computer calculation that the series converges to $\frac{p}{(1-p)^2}$, but I can't see why. Thanks in ...
- 147