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Let $$S = \sum^{\infty}_{n=1} a(n)$$ be an infinite series such that the nth partial sum is given by: $$S(n) = \frac{n - 1}{n + 1}$$

since $$ a(n)=S(n)-S(n-1)=\frac{2}{n(n+1)} $$

Now, $S(1)=a(1)\Leftrightarrow 0=1.$ Where is the mistake?

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    $\begingroup$ $$ a(1) = \frac{2}{1\times 2} = 1 = S(1) - S(0) = 0 - \left(\frac{0-1}{0+1}\right) = 1 $$ $\endgroup$ – Anonymous Dec 11 '16 at 21:18
  • $\begingroup$ It is not mandatory that a induction holds, also base step have to hold, but see the answers $\endgroup$ – GameDeveloper Dec 12 '16 at 12:00
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Good example of why you have to be careful with bounds :) Let me fix your post:

Let $$S = \sum^{\infty}_{n=1} a(n)$$ be an infinite series such that the nth partial sum is given by: $$S(n) = \frac{n - 1}{n + 1}$$

for all $\boldsymbol{n \ge 1}$, and $\boldsymbol{S(0) = 0}$. It follows that

$$ a(n)=S(n)-S(n-1)=\frac{2}{n(n+1)} $$

for all $\boldsymbol{n \ge 2}$.

Now, $S(1)=a(1)\Leftrightarrow 0=1.$

But now this falls apart, because the previous statement only held for $n \ge 2$.

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The mistake is in saying that the $n$'th partial sum is $(n-1)/(n+1)$, with the implication that this applies to $n=0$ as well as $n=1$.

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Depending on your definition there are two possible problems:
If S(0) exists: Then $S(0)$ has to definitions as $S(0)=\sum_{n=1}^0a(n)=0$ and $S(0)=\frac{0-1}{0+1}=-1$
If $S(0)$ doesn't exists: Then $S(1)$ is your first term and $a(1)=S(1)-S(0)$ isn't a valid formula for $a(1)$

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When

$S_n-S_{n-1}=f(n)-f(n-1)$,

you also have

$S_n-S_{n-1}=(f(n)+C)-(f(n-1)+C)$

for any constant $C$.

So you can only guarantee $S_n=f(n)+K$ for some constant $K$. If $K$ happens not to be zero for your choice of $f(n)$ and you are caught unawares, you get the discrepancy reported in the question. You have to check the first term specifically to nail down the correct value if $K$.

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