infinity sum of the fractional

Question

Can anyone explain how to simplify $ \frac{2}{3} + \frac{6}{9} + \frac{12}{27} + \frac{20}{81} + \frac{30}{243} + . . . $

I have no any idea since i dont have pattern i can't do it with integral or maybe differentation So can anyone help?

Answer 1

$$S=\sum_{k=1}^{\infty}\frac{k(k+1)}{3^k};\quad\frac13S=\sum_{k=1}^{\infty}\frac{k(k+1)}{3^{k+1}}=\sum_{k=2}^{\infty}\frac{(k-1)k}{3^k}$$ Subtracting: $$\frac23S=\frac23+\sum_{k=2}^{\infty}\frac{2k}{3^k}\implies S=1+3\sum_{k=2}^{\infty}\frac{k}{3^k};\quad S=\sum_{k=1}^{\infty}\frac{k}{3^{k-1}}$$ Similarly: $$\frac23S=1+\sum_{k=2}^{\infty}\frac1{3^{k-1}}=1+\sum_{k=1}^{\infty}\frac1{3^k}=1+\frac12=\frac32$$ So: $$\huge \rm S=\frac94$$

Answer 2

The numerators are $2,6,12,20,30,\ldots$, half of which is $1,3,6,10,15,\ldots$, a very familiar sequence: the $n$-th term here is $\sum_{k=1}^nk=\frac12n(n+1)$, so doubling it yields $n(n+1)$.

The denominators are clearly consecutive powers of $3$.

Thus, if we start the indexing at $1$, so that $a_1=\frac23$ and $a_2=\frac69$, we’re looking at

$$a_n=\frac{n(n+1)}{3^n}\;,$$

and the series is

$$\sum_{n\ge 1}a_n=\sum_{n\ge 1}\frac{n(n+1)}{3^n}\;.$$

Consider the series $$\sum_{n\ge 1}n(n+1)x^n\;,$$

with $x=\frac13$. Since you mention differentiation, you might try starting with a simpler series and differentiating it a time or two.

Answer 3

Try $$ \frac{i\cdot(i+1)}{3^i} $$ for $i=1,2,3,...$.

Answer 4

Following Brian M. Scott, consider $$S=\sum_{n=1}^m n(n+1)x^n$$ What you know is that starting with $x^n$, the first derivative is $n x^{n-1}$, the second derivative is $n (n-1)x^{n-2}$ and so on.

So start with $n(n+1)=An(n-1)+Bn$ and identify $A$ and $B$. So, once done, you have $$S=\sum_{n=1}^m n(n+1)x^n=\sum_{n=1}^m \Big( An(n-1)+Bn\Big)x^n=Ax^2\sum_{n=1}^m n(n-1)x^{n-2}+Bx\sum_{n=1}^m nx^{n-1}$$

I am sure that you can take from here.