MHB Help with understanding this series

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The discussion focuses on understanding a series and the challenges faced with applying the integral test. A participant suggests using the limit comparison test, specifically evaluating the ratio of consecutive terms. Another contributor demonstrates that the series in question converges by comparing it to a known convergent geometric series. The conclusion is that since the series is bounded by a convergent series, it also converges. This highlights the effectiveness of the comparison test in determining convergence.
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Can anyone help with this problem. I've tried integral test but seems to be too complicated.View attachment 9622
 

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Linus12351 said:
Can anyone help with this problem. I've tried integral test but seems to be too complicated.
Have you tried looking at [math]\lim_{n \to \infty} \dfrac{a_{n + 1}}{a_n}[/math]?

-Dan
 
Linus12351 said:
Can anyone help with this problem. I've tried integral test but seems to be too complicated.

Easy,

$\displaystyle 0 \leq \sum{\frac{1}{5^{n-1} + 1}} < \sum{\frac{1}{5^{n-1}}} = \sum{ \left( \frac{1}{5} \right) ^{n-1} }$

Since your positive term series is less than a convergent geometric series, your series converges by comparison.
 

Thread 'Proving that convexity implies second order derivative being positive'

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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