MHB Solving Series Limit Problem: Find Convergence/Divergence

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The discussion centers on determining the convergence or divergence of the infinite series $$\sum_{k = 1}^{\infty} {(\frac{e }{3})}^{k}$$. Participants suggest using the formula for infinite geometric series, which converges if the common ratio is between -1 and 1. In this case, since $$\frac{e}{3}$$ is less than 1, the series converges. The convergence can be confirmed by applying the geometric series test. The series converges to a specific value based on this analysis.
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I have this limit:

$$\sum_{k = 1}^{\infty} {(\frac{e }{3})}^{k}$$

Which method can I use to find if it converges or diverges?
 
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tmt said:
I have this limit:

$$\sum_{k = 1}^{\infty} {(\frac{e }{3})}^{k}$$

Which method can I use to find if it converges or diverges?

Look up infinite geometric series.
 

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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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