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Test the convergence

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

if \(x>1\)

Note by Tanishq Varshney 2 years, 2 months ago

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@Brian Charlesworth sir, @Satyajit Mohanty @Michael Mendrin sir

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Well, is it convergent?

dont know , was given homework

@Tanishq Varshney – Where do you study? Undergraduate or High School or Senior Secondary (+2) ?

@Satyajit Mohanty – under graduate, btw thanks for helping, are u sure it diverges

@Tanishq Varshney – 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 \(|r| \geq 1\).

@Satyajit Mohanty – one more doubt is \(\displaystyle \sum_{r=1}^{\infty} r\) convergant or divergant. Becoz ramanujan proved this equal to -1/12

@Tanishq Varshney – \(\displaystyle \sum_{r=1}^\infty r \) is divergent. But again, Analytic Continuation changes the game and we find it as \(-\frac{1}{12}\).

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## Comments

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TopNewest@Brian Charlesworth sir, @Satyajit Mohanty @Michael Mendrin sir

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Well, is it convergent?

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dont know , was given homework

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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 \(|r| \geq 1\).

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Analytic Continuation changes the game and we find it as \(-\frac{1}{12}\).

\(\displaystyle \sum_{r=1}^\infty r \) is divergent. But again,Log in to reply