A coin has probability $p$ of landing heads, and $q=1-p$ of landing tails. Assume that $p\ne 0$.
Let $X$ be the total number of tosses until you get a head. Then $X$ has precisely the geometric distribution that you described. One can, as you did, get an expression for $E(X)$ as an infinite series. In fact, it turns out $E(X)=1/p$. But we need neither the series nor its sum to prove the result that is asked for.
Suppose that we are given that $X>1$. This means that our first toss was a tail. Let $Y$ be the additional number of tosses that we must wait for a head. The coin does not remember that the first toss was a tail, so $Y$ has the same distribution, and therefore the same mean, as $X$. In symbols, $E(Y)=E(X)$.
But the total number of tosses, given that $X>1$, is $1+Y$. The $1$ is for the "wasted" first toss. Thus
Comment: If you prove the result using the infinite series, you know that the result is true. If you do it more conceptually, you know why the result is true.