4

$$\sum_{k=1}^{\infty} \left(\frac{1}{k+3}-\frac{1}{k+4}\right)$$

It is a telescopic series , so I write it as:

$a_n=\frac{1}{4}-\frac{1}{k+4}$ taking the limit $\lim_{k\to \infty}\frac{1}{4}-\frac{1}{k+4}=\frac{1}{4}$

Is it legit?

Leucippus
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gbox
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  • This is the right idea, but you might want to justify (e.g., by induction) that your formula for the $n$th partial sum is correct. – Travis Willse Mar 15 '16 at 16:42
  • almost, I think you mean $a_k$ instead of $a_n$ or an $n$ in the expression for it – Jens Renders Mar 15 '16 at 16:42
  • @Travis I can say that the general expression is $\frac{1}{4}-\frac{1}{n}$ or $\frac{1}{4}-\frac{1}{n+1}$ ? – gbox Mar 15 '16 at 16:45
  • Yes, depending on the upper index for your partial sum. – Travis Willse Mar 15 '16 at 16:48
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    @gbox up to using the right index, I understand what you wrote as a shorthand for $$a_k:=\sum_{h=1}^k\frac{1}{h+3}-\frac{1}{h+4}=\frac14-\frac{1}{k+4}$$ so you should use $k+4$. Of course, if you meant $b_k:=\sum_{h=1}^{k-4}\text{etc...}=\frac14-\frac1k$ you can use the latter, since shifting by a couple of spaces does not change the limit. –  Mar 15 '16 at 16:51

1 Answers1

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Yes, you are correct. The sum telescopes, and only the first term $1/4$, survives in the limit.

Aside: I might add that your notation $a_k$ for the $k$th partial sum is an interesting notational choice, as so frequently by convention we think of the original sum as $\sum a_n$. Of course, this is merely convention. But when I teach telescoping series to my students, I usually use the notation $\Sigma_k$ or $S_k$ for the $k$th partial sum.

davidlowryduda
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