Changing starting point of a sum to another one.

When you’re just starting to deal with summations it may be easier to see what’s going on here by making an actual substitution for the index variable. To avoid visual clutter I’ll let $a=\sin 1$. Suppose that I let $k=n-1$; then $n=k+1$, and we have

$$\sum_{n=1}^\infty a^n=\sum_{k+1=1}^\infty a^{k+1}=\sum_{k=0}^\infty a^{k+1}\;,$$

since as $k+1$ runs over all positive integers, $k$ itself clearly runs over all non-negative integers. Clearly $a^{k+1}=a\cdot a^k$, so

$$\sum_{k=0}^\infty a^{k+1}=\sum_{k=0}^\infty\left(a\cdot a^k\right)=a\sum_{k=0}^\infty a^k\;,$$

where the last step is justified by the fact that we can always pull out of a summation a factor that does not depend on the index variable (i.e., one that really is the same in every term).

Now just apply the formula for the sum of a geometric series to $\sum_{k=0}^\infty a^k$ to get

$$a\sum_{k=0}^\infty a^k=a\cdot\frac1{1-a}=\frac{a}{1-a}\;.$$

Putting the pieces together, we have

$$\sum_{n=1}^\infty a^n=\frac{a}{1-a}\;.$$

In effect the whole computation is just noticing that

$$\begin{align*} a+a^2+a^3+a^4+\ldots&=a\cdot1+a\cdot a+a\cdot a^2+a\cdot a^3+\ldots\\ &=a\left(1+a+a^2+a^3+\ldots\right)\\ &=a\cdot\frac1{1-a}\\ &=\frac{a}{1-a}\;. \end{align*}$$

Note that at beginning, after I established that

$$\sum_{n=1}^\infty a^n=\sum_{k=0}^\infty a^{k+1}\;,$$

I could have renamed the index variable back to $n$ and simply written

$$\sum_{n=1}^\infty a^n=\sum_{n=0}^\infty a^{n+1}\;.$$

With practice one can make this sort of shift automatically, without going through the substitution and renaming at all.

When you change the starting point of $$\sum_{n=1}^\infty (\sin 1)^n$$

to 0, the sum would become

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

which can be written as

$$\sum_{n=0}^\infty (\sin 1)^{n} (\sin 1)^1$$