I am trying to simplify the following expression

$$ \left(\sum_{n=0}^{\infty} t^{2n}\right)^2 $$

I tried finding out a pattern for the coefficients by writing just the first three terms and squaring, i.e

$$ (1+t^2+t^4)^2=1+2t^2+3t^4+2t^6+t^8 $$ and I notice that the more and more terms I square, there seems to be a pattern,i.e for 3 terms, the coefficients are $ 1,2,3,2,1$ for 4 terms it is $1,2,3,4,3,2,1$ and so on. But could someone please tell me how I can get a closed form summation expression with this?

  • $\begingroup$ Solve $\sum_{n=0}^\infty r^n$ where $r=t^2$ and the sum is geometric series, then square the result. $\endgroup$ – Simply Beautiful Art Oct 17 '16 at 1:28
  • $\begingroup$ Well, the closed form is $\frac{1}{(1-t^2)^2}$. Do you want to know the coefficients? $\endgroup$ – Thomas Andrews Oct 17 '16 at 1:28
  • $\begingroup$ @ThomasAndrews Following up with the OP's expansions, it is clearly $1+2t^2+3t^3+\dots$ $\endgroup$ – Simply Beautiful Art Oct 17 '16 at 1:29
  • $\begingroup$ Well, $1+2t^2+3t^{\mathbf 4}+\cdots$. @SimpleArt $\endgroup$ – Thomas Andrews Oct 17 '16 at 1:30
  • $\begingroup$ @ThomasAndrews True, forget that inside $t^2$ $\endgroup$ – Simply Beautiful Art Oct 17 '16 at 13:37

Well, if you really insist on writing out $(\sum t^{2n})^2$ as a new summation $\sum a_n t^n$, rather than simply using the other answers for a closed form expression, it is going to really on the binomial theorem / collecting terms in the appropriate manner.

Looking at $$\left(\sum_{n=0}^\infty t^{2n} \right)^2 = (1 + t^2 + t^4 + \cdots)(1+t^2 + t^4 + \cdots)$$ we want to observe how many ways there are to produce terms of the appropriate exponent:

  1. Constant terms: we can only pick $1$ from the first terms and $1$ from the second, i.e. $1\cdot 1$.

  2. $t^{2k+1}$ terms: No possibilities--both expressions only include $t$ to even powers, so no multiplication will give odd exponents.

  3. $t^2 terms$: we can pick $1\cdot t^2$ or $t^2 \cdot 1$, for a coefficient of 2

  4. $t^4 terms$: we can pick $1 \cdot t^4$, $t^2\cdot t^2$, or $t^4 \cdot 1$. Thus the coefficient is 3.

  5. $t^6 terms$: we can pick $1 \cdot t^6$, $t^2 \cdot t^4$, $t^4 \cdot t^2$, or $t^6 \cdot 1$. Coefficient is 4.

  6. $t^{2n} terms$: we can pick $1 \cdot t^{2n}$, $t^2 \cdot t^{2n-2}$, $\dots$, $t^{2n} \cdot 1$. There are $n+1$ of these.

Thus the general formula would be $$\left(\sum_{n=0}^\infty t^{2n}\right)^2 = \sum_{n=0}^\infty (n+1)t^{2n}$$

Note that if we were presented with $(\sum a_n t^{2n})^2$ instead, the answer would be much messier and would involve a double sum: $(\sum a_n t^{2n} )^2= \sum_n \sum_j a_j t^{2n}$


$${\frac {1}{1-t}}=\sum _{n=0}^{\infty }t^{n}\quad {\text{ for }}|t|<1$$ let $t\rightarrow t^2$ $${\frac {1}{1-t^2}}=\sum _{n=0}^{\infty }t^{2n}$$ $$({\frac {1}{1-t^2}})^2=(\sum _{n=0}^{\infty }t^{2n})^2$$ we have $${\frac {1}{(1-t^2)^{2}}}=\sum _{n=1}^{\infty }nt^{2n-2}$$ so $$(\sum _{n=0}^{\infty }t^{2n})^2=\sum _{n=1}^{\infty }nt^{2n-2}=\sum _{n=0}^{\infty }(n+1)t^{2n}$$

  • $\begingroup$ Thank you, but I was looking for a summation expression of the form $\endgroup$ – Math_Enthusiast Oct 17 '16 at 1:35
  • $\begingroup$ $ \sum_{n=0}^{\infty} a_{n} $ $\endgroup$ – Math_Enthusiast Oct 17 '16 at 1:37
  • $\begingroup$ Is this right? ${\frac {1}{(1-t^2)^{2}}}=\sum _{n=1}^{\infty }nt^{2n-1}$ $\endgroup$ – gowrath Oct 17 '16 at 1:59
  • $\begingroup$ @gowrath thanks , I have corrected it. $\endgroup$ – E.H.E Oct 17 '16 at 2:17

\begin{align} \left(\sum_{n=0}^\infty t^{2n}\right)^2&=\left(\sum_{j=0}^\infty t^{2j}\right)\left(\sum_{k=0}^\infty t^{2k}\right) \\&=\sum_{j=0}^\infty \sum_{k=0}^\infty t^{2j+2k} \\&=\sum_{n=0}^\infty \text{(# of ways of writing }2n\text{ as the sum of two non-negative even integers)}\cdot t^{2n} \\&=\sum_{n=0}^\infty \text{(# of ways of writing }n\text{ as the sum of two non-negative integers)}\cdot t^{2n} \\&=\sum_{n=0}^\infty (n+1)\, t^{2n}. \end{align}


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