I am trying to prove the properties of the telescoping series via an exercise in Tao's analysis text. The exercise, with the full proposition filled in, is:

Let $(a_n)_{n=0}^{\infty}$ be a sequence of real numbers which converge to $0$, i.e., $\lim\limits_{n \to \infty} a_n = 0$. Then the series $\sum\limits_{n=0}^{\infty} (a_n - a_{n+1})$ converges to $a_0$. How does the proposition change if we assume that $a_n$ does not converge to zero, but instead converges to some other real number, $L$?

Here is my attempt at a proof.

Per the hint in Tao, we guess a formula for the $n$th partial sum, $S_n$, and prove it by induction.

Lemma: For all $n \in \mathbb{N}$, $S_n = a_0 - a_{n+1}$.

Base Case: $S_0 = \sum\limits_{n=0}^{0} (a_n - a_{n+1}) = a_0 -a_1$.

Induction Hypothesis: Assume $S_k = \sum\limits_{n=0}^{k} = a_0 - a_{k+1}$.

Induction Step: We prove the stateent for $n = k + 1$. \begin{align*} s_{k+1} & =\sum\limits_{n=0}^{k+1} (a_n - a_{n+1}) \\ & = \sum\limits_{n=0}^k (a_n - a_{n+1}) + (a_ {k+1} - a_{k+2}) \\ & = s_k + (a_{k+1} - a_{k+2}) \\ & = a_0 - a_{k+1} + a_{k+2} - a_{k+2} \\ & = a_0 - a_{k+2} \end{align*} This closes the induction.

We have: \begin{align*} \lim s_n =\lim\limits_{n \to \infty} [a_0 - a_{n+2}] = \lim\limits_{n \to \infty} a_0 - \lim\limits_{n \to \infty} a_{n+2} = a_0 - 0 = a_0. \end{align*} And since the series converges to the same limit as the sequence of partial sums, we conclude that the series also converges to $a_0$.

If $a_n$ converges to $L$, then $\lim\limits_{n \to \infty} a_{n+2} = L$, so the sequence of partial sums would converge to $a_0 - L$, and thus the series would converge to $a_0 - L$.

How does this look? Any helpful comments would be greatly appreciated.

  • 1
    $\begingroup$ That looks fine. $\endgroup$ – Lee Mosher Jun 17 '19 at 17:00

If $(b_n)_{n=0}^\infty$ is a sequence of real numbers which converge to $L$ then we can just let $a_n=b_n-L$ such that $a_n$ converges to $0$ hence $$\sum_{n=0}^\infty (b_n-b_{n+1})=\sum_{n=0}^\infty (a_n+L-(a_{n+1}+L))=\sum_{n=0}^\infty (a_n-a_{n+1})=a_0=b_0-L$$

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