Why $$\sum_{n=0}^{\infty}(n+1)5^nx^n=\frac{1}{(1-5x)^2}?$$

I know that $\sum_{n=0}^{\infty}x^n=\dfrac{1}{1-x}$, so by the same token, $\sum_{n=0}^{\infty}5^nx^n=\dfrac{1}{1-5x}$.

Thus $$ \left(\frac{1}{1-5x}\right)^2=\frac{1}{(1-5x)^2} = \left(\sum_{n=0}^{\infty}5^nx^n\right)^2. $$

But why is $\big(\sum_{n=0}^{\infty}5^nx^n\big)^2=\sum_{n=0}^{\infty}(n+1)5^nx^n$?

Assuming $x$ is small enough so that the sum converges.

  • 1
    $\begingroup$ See this. Note your series can be written as $\sum_{n=0}^\infty (n+1)(5x)^n$. $\endgroup$ – David Mitra Feb 5 '14 at 15:56

Note that $$ \frac{1}{1-x}=\sum_{n=0}^\infty x^n, $$ and thus $$ \frac{1}{(1-x)^2}=\left(\frac{1}{1-x}\right)'=\sum_{n=1}^\infty nx^{n-1}=\sum_{n=0}^\infty (n+1)x^{n}, $$ and hence $$ \frac{1}{(1-5x)^2}=\sum_{n=0}^\infty (n+1)(5x)^{n}. $$


What happens when you differentiate $\frac{1}{1-5x}$? How does this relate to your series?

  • $\begingroup$ I see now! Thank you. $\endgroup$ – Oria Gruber Feb 5 '14 at 15:58

Although the method via differentiating the known power series is what I'd recommend, here's a direct argument why $(\sum_{n=0}^{\infty}5^nx^n)^2=\sum_{n=0}^{\infty}(n+1)5^nx^n$ or more simply why $$(\sum_{k=0}^{\infty}x^k)^2=\sum_{n=0}^{\infty}(n+1)x^n.$$

The left-hand-side is $$(x^0 + x^1 + x^2 + \cdots + x^k + \cdots) \times (x^0 + x^1 + x^2 + \cdots + x^l + \cdots),$$ so the coefficient of $x^n$ in the LHS is the number of pairs $(k, l)$ with $k + l = n$. This is $n+1$, as there are $n+1$ pairs $(0,n), (1,n-1), \dots, (n,0)$. This is the same as the coefficient of $x^n$ in the RHS.

We can also give it a combinatorial interpretation: $F(x) = \sum_{n=0}^{\infty} x^n$ is the generating function for the nonnegative integers, and so $F(x)^2$ is the generating function for the number of ways of writing an integer as the sum of two nonnegative integers, which is $n+1$ (by the argument above).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.