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Why can't this series $$\sum_{n=1}^{\infty}\frac1{n^n}$$ be looked as a Geometric Series with $r=\frac1{n}$?

I am looking for answers other than "$r$ is supposed to be a fixed real number from the definition of geometric series".

Thank you.

gymbvghjkgkjkhgfkl
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DMH16
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  • No, in this case $r = \frac{(n - 1)^{n - 1}}{n^n}$ and similarly we can call any raw a Geometric Series. – Smylic Mar 04 '17 at 20:36
  • @Smylic how did you get such value for $r$? – DMH16 Mar 04 '17 at 20:38
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    So, in short, your question is "why is this thing which does not satisfy the definition of a geometric series [which would require $r$ to be a constant] cannot be said to be a geometric series -- and I am looking for answers other than 'because it does not satisfy the definition of a geometric series'"? – Clement C. Mar 04 '17 at 20:38
  • DMH16, I divided summand by previous one. – Smylic Mar 04 '17 at 20:39
  • @ClementC. Yes. – DMH16 Mar 04 '17 at 20:40
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    I don't mean to be mean, but the comment above was supposed to highlight the hopeless character of such a question. It is not a geometric series because it does not satisfy the definition of a geometric series. – Clement C. Mar 04 '17 at 20:41
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    @DMH16 But that's all there is to it. "Why can't we see the a trapezium as a special case of square?" Well, because it is not a square. And the answer to the (not really) deeper question "Why is there no trick that allows me to work on it as if it were a square?" is: because it is not a square and today you've been unlucky not to find the only case where using a wrong method yields a numerically exact result. –  Mar 04 '17 at 20:46
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    The ratio between successive terms is not constant. Therefore the series is not geometric. – Henricus V. Mar 04 '17 at 21:21

4 Answers4

9

Sorry to say, but the answer is "$r$ is supposed to be a fixed real number from the definition of geometric series".

3

The definition of geometric series is that the ratio of $\dfrac{a_{n+1}}{a_n}$ must be a non-zero constant real number.

Nosrati
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2

You can call anything anyhow to create chaos, but calling this raw a geometric series doesn't give you any profit, because all formulae concerning true geometric series wouldn't give you right result.

Smylic
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1

A geometric series with $r=\frac{1}{n}$ and $a=\frac{1}{n}$ would be

$$ \left(\frac{1}{n}\right)+\left(\frac{1}{n}\right)^2+\left(\frac{1}{n}\right)^3+\cdots=\sum_{k=1}^\infty\left(\frac{1}{n}\right)^k$$

and

$$ \sum_{k=1}^\infty\left(\frac{1}{n}\right)^k\ne\sum_{n=1}^\infty\left(\frac{1}{n}\right)^n $$

John Wayland Bales
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