= Boundary limits of telescoping series
{title2=$\sum_{k=1}^{\infty}(a_k-a_{k-1})=L-a_0\quad\text{if }a_n\to L$}
For a real <sequence> $(a_n)$, the <partial sum> of $\sum_{k=1}^{\infty}(a_k-a_{k-1})$ is exactly $a_n-a_0$. Consequently it is a <convergent series> if and only if $a_n$ has a finite <limit> $L$, and its <sum> is $L-a_0$. Cancellation takes place inside each finite <partial sum>; taking a <limit> is a separate step.<Calculus Volume 2 by Gilbert Strang and others/5.2 Infinite Series>[Calculus Volume 2, §5.2 Telescoping Series: general boundary sequence, finite partial sums and finite-limit convergence criterion]{ref}
The <partial fraction telescoping> example has $S_N=N/(3N+1)$ and therefore <limit> $1/3$. In contrast, let $a_n=(-1)^n$. Its first entry is $a_0=1$, so $S_N=(-1)^N-1$. The <even> <partial sums> are zero and the <odd> <partial sums> are negative two. Both finite cancellations are correct, but the <sequence> has no single <limit>, giving a <divergent series>.
\Image[/telescoping-series-boundary-convergence.png]
{title=Exact cancellation can leave a convergent or an oscillating <boundary term>}
{description=Original calculated <partial sums>. Top: $S_N=N/(3N+1)$ approaches $1/3$, with known <series remainder> $1/[3(3N+1)]$. Bottom: $S_N=(-1)^N-1$ alternates between zero and negative two. The dots are finite <partial sums>, not observed data.}
It is also essential to keep the two contributions paired. For example, $\sum_{k=1}^{\infty}(1/k-1/(k+1))$ is a <convergent series> with <sum> one, as its finite <partial sum> is $1-1/(N+1)$. Writing it instead as $\sum_{k=1}^{\infty}1/k-\sum_{k=1}^{\infty}1/(k+1)$ asks to subtract two <divergent series>. The permitted finite cancellation does not define that separate subtraction. Working through <partial sums> keeps the indices aligned and the operation meaningful.
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