4
votes
Accepted
Find the maximal region in $\mathbb C$ in which the following series $\sum\limits_{n=1}^\infty \frac{(-1)^{3n}n-n^2}{n+4}z^n$ is uniformly convergent
$$\lim_{n\to\infty}\sup\left|\frac{\frac{(-1)^{3(n+1)}(n+1)-(n+1)^2}{(n+1)+4}}{\frac{(-1)^{3n}n-n^2}{n+4}}\right|=\lim_{n\to\infty}\left|\frac{(n+4)((-1)^{3(n+1)}(n+1)-(n+1)^2)}{(n+5)((-1)^{3n}n-n^2)}\...
- 2,338
2
votes
Find the maximal region in $\mathbb C$ in which the following series $\sum\limits_{n=1}^\infty \frac{(-1)^{3n}n-n^2}{n+4}z^n$ is uniformly convergent
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2((-1)^{3(n+1)}/(n+1)^2-1)}{(n+1)(1+4/(n+1))}\cdot\frac{n(1+4/n)}{n^2((-1)^{3n}/n^2-1)}$$
for $n\to \infty $ we can see that the oscillating terms go to zero and we ...
- 2,056
2
votes
How to obtain asymptotic form of the integral $\int_0^\infty \frac{dx}{x^2 + 1} e^{-\alpha \sqrt{x^2 + 1}}$ for small and large values of $\alpha$?
This is too long for a comment.
Starting from @Travis Willse's nice answer, the expansion for small $\alpha$ could write
$$I(\alpha)= \frac \pi 2+ (A-1)\alpha+\frac{3 A-4}{36} \alpha ^3 +\frac{ 10
...
- 274k
1
vote
show that $ \frac{(n+1)^{r+1} - (n+1)}{(r+1)!} = \frac{S_1}{r!1!} + \frac{S_2}{(r-1)!2!} + \dots + \frac{S_r}{1!r!} $
$$
(n+1)^{r+1} -1 = \sum_{x=1}^n [(1+x)^{r+1} - x^{r+1}] = \sum_{x=1}^n \sum_{k=0}^r \binom{r+1}{k} x^k = \sum_{k=0}^r \binom{r+1}{k} S_k.
$$
Moving the $k=0$ term to the LHS and dividing by $(r+1)!$ ...
- 3,583
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