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While doing some homework ive came across problems were the solution involves changing the $\sum$ starting point from lets say $n=1$ to $n=0$ and in this process Ive noticed they will add a constant or so and I dont remember seeing this in class, and Im not quite sure why its done.

This is an example $$\sum_{n=1}^{\infty}(\sin1)^n$$ It then goes and rewrites it in terms of $n=0$ and does the following $$\sin(1)\sum_{n=0}^{\infty}\sin(1)$$ And they still use the extra value in the geometric series setup $$\frac{\sin1}{1-\sin1}$$ Im not sure if my question is clear enough but I would like to know how the extra value of the $\sin1$ is added in the second $\sum$ while only changing the starting point from $n=1 \to n=0$

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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.

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  • $\begingroup$ Thank you, it makes a lot more sense now. $\endgroup$ – Carlos V Sep 5 '16 at 20:38
  • $\begingroup$ @Carlos: You’re welcome. $\endgroup$ – Brian M. Scott Sep 5 '16 at 20:39
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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

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

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    $\begingroup$ I think that the OP needs a deeper explanation than this "magic". $\endgroup$ – Crostul Sep 5 '16 at 16:24
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\begin{align} \sum_{n=1}^\infty r^n & = r + r^2 + r^3 + r^4 + \cdots \\[10pt] \sum_{n=0}^\infty r^{n+1} & = r + r^2 + r^3 + r^4 + \cdots \end{align} Think about the meaning of $\displaystyle \sum_{n=1}^\infty$ and of $\displaystyle \sum_{n=0}^\infty$ and see if that tells you that the two expressions involving $\displaystyle\sum$ above both yield the expressions you see to the right of the "equals" signs above, and note that those two expressions are the identical.

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