Test the convergence

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

if \(x>1\)

Test the convergence

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

if \(x>1\)

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## Comments

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TopNewest@Brian Charlesworth sir, @Satyajit Mohanty @Michael Mendrin sir – Tanishq Varshney · 1 year, 9 months ago

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– Satyajit Mohanty · 1 year, 9 months ago

Well, is it convergent?Log in to reply

– Tanishq Varshney · 1 year, 9 months ago

dont know , was given homeworkLog in to reply

– Satyajit Mohanty · 1 year, 9 months ago

Where do you study? Undergraduate or High School or Senior Secondary (+2) ?Log in to reply

– Tanishq Varshney · 1 year, 9 months ago

under graduate, btw thanks for helping, are u sure it divergesLog in to reply

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 · 1 year, 9 months ago

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– Tanishq Varshney · 1 year, 9 months ago

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

Analytic Continuation changes the game and we find it as \(-\frac{1}{12}\). – Satyajit Mohanty · 1 year, 9 months ago

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