MHB Interval of Convergence: Find Series

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To find the interval of convergence for the series, the ratio test is recommended. For the first series, evaluate the limit of the ratio of consecutive terms and solve for x to determine convergence. The second series also requires applying the ratio test to find the appropriate interval. The third series can be analyzed similarly by assessing the limit of the ratio of terms. Each series will yield a specific interval where convergence occurs based on these calculations.
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How would I find the interval of convergence for the following series:

i) $$\sum \frac{(x+2)^n}{n^2}$$

ii) $$\sum \frac{(-1)^kk^3}{3^k}(x-1)^{k+1}$$

iii) $$\sum (1+\frac{1}{n})^nx^n$$

Reason for edit: My second series was not displaying properly
 
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I would ask that you edit the second expression so that it renders as you want, and then show us what you have tried so we know where you are stuck and can offer better help. :D
 
MarkFL said:
I would ask that you edit the second expression so that it renders as you want, and then show us what you have tried so we know where you are stuck and can offer better help. :D
Sorry about that. Done. btw big Rush fan myself :)
 
Anewk said:
How would I find the interval of convergence for the following series:

i) $$\sum \frac{(x+2)^n}{n^2}$$

ii) $$\sum \frac{(-1)^kk^3}{3^k}(x-1)^{k+1}$$

iii) $$\sum (1+\frac{1}{n})^nx^n$$

Reason for edit: My second series was not displaying properly

The ratio test states that for any series $\displaystyle \begin{align*} \sum_{\textrm{all }n} a_n \end{align*}$ is convergent if $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}$ and divergent where $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1 \end{align*}$. The test is inconclusive if the limit is 1.

So in each of these, evaluate $\displaystyle \begin{align*} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$, evaluate its infinite limit, set it less than 1, and solve for x.
 
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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