Rewriting an infinite sum

Question

Rewrite the given expression as a sum whose generic term involves x^n:

$$ x\cdot\sum_{n=1}^{\infty}(n a_n x^{n-1}) + \sum_{k=0}^{\infty}(a_k x^{k} ) $$

I get a sum starting at one:

$$ \sum_{n=1}^{\infty}(n a_n x^{n}) + \sum_{n=1}^{\infty}(a_n x^{n} ) = \sum_{n=1}^{\infty}(n+1)a_n x^{n} $$

Whereas the answer is a similar sum starting at zero.

$$ \sum_{n=0}^{\infty}(n+1)a_n x^{n} $$

Where am I going wrong?

I figured that k could be viewed as a dummy variable and exchanged it for one involving n instead. Is this where my error could be? — V.Vocor, Oct 10 '14 at 20:28
The second sum in the second displayed identity should start at $n=0$ instead of $n=1$. — Did, Oct 10 '14 at 20:37
@V.Vocor: Yes, $k$ is a dummy variable, but you don't get to change the lower limit from $k=0$ to $n=1$ just because you rename the variable. The sum should still be from $n=0$. — hmakholm left over Monica, Oct 11 '14 at 13:07

score 1 · Accepted Answer · answered Oct 10 '14 at 20:35

My advice is to write out the first several terms of the expression: $$x \sum_{n=1}^\infty na_n x^{n-1} + \sum_{k=0}^\infty a_k x^k = x(a_1 x^0 + 2a_2 x^1 + 3a_3 x^2 + \cdots) + (a_0 x^0 + a_1 x^1 + a_2 x^2 + \cdots).$$ Now equating like powers of $x$ gives the result $$a_0 x^0 + 2a_1 x^1 + 3a_2 x^2 + \cdots.$$ This suggests that the general term should have the form $(n+1)a_n x^n$, for $n = 0, 1, 2, \ldots$. All that remains is to formalize this by expressing the summands with corresponding indices of summation.

This is a helpful answer as well. Thanks! – V.Vocor Oct 10 '14 at 20:42 — V.Vocor, Oct 10 '14 at 20:42

Rewriting an infinite sum

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