Test the convergence

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

if $$x>1$$

Note by Tanishq Varshney
3 years, 1 month ago

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- 3 years, 1 month ago

Well, is it convergent?

- 3 years, 1 month ago

dont know , was given homework

- 3 years, 1 month ago

Where do you study? Undergraduate or High School or Senior Secondary (+2) ?

- 3 years, 1 month ago

under graduate, btw thanks for helping, are u sure it diverges

- 3 years, 1 month ago

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 $$|r| \geq 1$$.

- 3 years, 1 month ago

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

- 3 years, 1 month ago

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

- 3 years, 1 month ago