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We know that $0 \times \infty$ is an indeterminate form. However, is it equivalent to $0 + 0 + 0 + \cdots$? If yes, why we do not consider $\displaystyle \sum_{n = 0}^\infty 0$ an indeterminate form?

--EDIT--

We can also write the sum for any constant $k$ from $0$ to $n$ as $k(n+1)$

So, $\displaystyle \sum_{n=0}^\infty 0 = 0 \times (\infty + 1)$ which is an IF.

Why does wolframalpha say that it is convergent?

Thank you,

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Use the definition of an infinite series, the limit of the sequence of partial sums:

$$\displaystyle \sum_{k=0}^\infty 0 = \lim_{N \to +\infty} \sum_{k = 0}^N 0$$

By the definition of limits at infinity, for every $\epsilon > 0$, choose any $M$ you want, it doesn't matter because the sequence is constant:

$$\displaystyle N > M \implies -\epsilon < \sum_{k = 0}^N 0 < \epsilon $$

As for your edit,

$$\displaystyle \sum_{k=0}^\infty 0 = \lim_{N \to +\infty} \sum_{k = 0}^N 0 = \lim_{N \to +\infty}0(N+1) = 0$$

Indeterminate form doesn't mean the limit doesn't exist, it means it may or may not exist, and here the series converges.

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  • $\begingroup$ Amazing! I do not know how I missed these basic ideas. Thank you so much. $\endgroup$ – user209213 Jan 20 '15 at 16:14
  • $\begingroup$ @user209213 They're only basic after you see them answered! Please accept my answer if you like it :) $\endgroup$ – GFauxPas Jan 20 '15 at 16:53
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Definition of infinite sums is in terms of limit of finite sums: namely the partial sums.

Now, adding finitely many $0's$ only gives you $0$. So, $S_n=0$ for all $n$.

The limit of a sequence whose terms are all $0$ is again $0$.

So, the series is convergent, and the sum is $0$.

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