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Let $a_1=1\ $ and $\ a_n=2-\frac 1n$ for $n\geq 2$. Then $$\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)$$ converges to?


I started by expanding the sum:

$\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)=\bigg(\frac{1}{a_1^2}-\frac{1}{a_2^2}\bigg)+\bigg(\frac{1}{a_2^2}-\frac{1}{a_3^2}\bigg)+\bigg(\frac{1}{a_3^2}-\frac{1}{a_4^2}\bigg)+\ldots$

$=\frac{1}{a_1^2}+\bigg(-\frac{1}{a_2^2}+\frac{1}{a_2^2}\bigg)+\bigg(-\frac{1}{a_3^2}+\frac{1}{a_3^2}\bigg)+\bigg(-\frac{1}{a_4^2}+\frac{1}{a_4^2}\bigg)+\ldots$

$=\frac{1}{a_1^2}$

$=1$


Now, the strange part is that the answer to this question is known to me as $0.75$. Is there something wrong with what I did? How does one arrive at $0.75$ as a solution?

PS: I am aware that infinite sums can behave in strange ways, and sometimes more than one solution may follow logically. Is this one such case?

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Let $$S_N=\sum_{n=1}^N\left(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\right)=\frac{1}{a_1^2}-\frac{1}{a_{N+1}^2}.$$ Since $a_1=1$ and $a_n\to 2$, we have $$\lim_{N\to\infty}S_N=1-\frac{1}{2^2}=\frac{3}{4}.$$

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Infinite sums can behave in strange ways

True!

Sometimes more than one solution may follow logically

Definitely not true. There is one answer in every case (as long as you allow "the sum diverges" as an answer). Sure, there are subtle errors which may make it appear that a sum has more than one value, but that's exactly what they are - errors.

To avoid these errors, don't manipulate infinite sums unless you have a conclusive proof that the manipulation is justified. If you don't, go back to the definition. In your case this says $$\eqalign{\sum_{n=1}^\infty\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg) &=\lim_{N\to\infty}\sum_{n=1}^N\bigg(\frac{1}{a_n^2}-\frac{1}{a_{n+1}^2}\bigg)\cr &=\lim_{N\to\infty}\biggl(\frac{1}{a_1^2}-\frac{1}{a_{N+1}^2}\biggr)\ .\cr}$$ See if you can finish the job from here.

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  • $\begingroup$ I stated that more than one solution may follow logically because of Grandi's series where 0, 1, and half are possible answers. Although, I suppose you could simply call the series divergent and completely avoid the controversy altogether. $\endgroup$ – s0ulr3aper07 Feb 4 at 3:23
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    $\begingroup$ Using standard definitions, Grandi's series is divergent and that's that. Using Cesàro summation you get $\frac12$, but this does not mean the series actually converges to $\frac12$. As far as I am aware, there has been absolutely no serious controversy on this question for well over a century. Well stated in Wikipedia: "the above manipulations [giving various 'sums'] do not consider what the sum of a series actually means". $\endgroup$ – David Feb 4 at 3:53
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$a_n$ does not approach zero, so the end term can not be disregarded.

In other words, the sum up to $n$ is $\frac1{a_1^2} -\frac1{a_n^2} \to 1-\frac14 =\frac34 $.

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