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# Anyone with answer

Test the convergence

$\large{\displaystyle \sum_{n=1}^{\infty} x^n}$

if $$x>1$$

Note by Tanishq Varshney
1 year, 9 months ago

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Well, is it convergent? · 1 year, 9 months ago

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dont know , was given homework · 1 year, 9 months ago

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Where do you study? Undergraduate or High School or Senior Secondary (+2) ? · 1 year, 9 months ago

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under graduate, btw thanks for helping, are u sure it diverges · 1 year, 9 months ago

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Well Hmm, yup it's divergent! But anything can happen due to analytic continuations :D

For example $$\zeta(-2n) = 0$$ $$\Rightarrow 1^{2n} + 2^{2n} + 3^{2n} + 4^{2n} + \ldots = 0$$ for $$n \in \mathbb {Z}^+$$

I just know that:

$$\large{\displaystyle \sum_{n=1}^{\infty} ar^n}$$ converges only when $$|r| < 1$$, $$a \neq 0$$. It diverges for $$|r| \geq 1$$. · 1 year, 9 months ago

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one more doubt is $$\displaystyle \sum_{r=1}^{\infty} r$$ convergant or divergant. Becoz ramanujan proved this equal to -1/12 · 1 year, 9 months ago

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$$\displaystyle \sum_{r=1}^\infty r$$ is divergent. But again, Analytic Continuation changes the game and we find it as $$-\frac{1}{12}$$. · 1 year, 9 months ago

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