# Homework Help: Alternating serries

1. Aug 23, 2009

### nameVoid

$$\sum_{n=1}^{\infty}\frac{(-10)^n}{n!}$$
im seeing the terms increasing here DIVERGENT
$$\sum_{n=1}^{\infty}\frac{10^n}{n!}$$
$$\lim_{n->\infty}\sqrt[n]{\frac{10^n}{n!}}>1$$
text is showing absolute convergence
$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2}$$
$$\lim_{n->\infty}\frac{n^2+3}{(2n-5)^2}\neq0$$
divergent
$$\frac{n^2+3}{(2n-5)^2}\leq1$$
$$\sum_{n=1}^{\infty}1 -> \infty$$
$$\lim_{n->\infty}{\frac{n^2+3}{(2n-5)^2}>0$$
divergent by limit comparison test

1. The problem statement, all variables and given/known data

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2. Aug 23, 2009

### zcd

You might want to try ratio test (for absolute convergence) in the case of factorials.
$$\lim_{n\to\infty}\left|\frac{10^{n+1}}{(n+1)n!}\cdot\frac{n!}{10^{n}}\right|=\lim_{n\to\infty}\left|\frac{10}{n+1}\right|=0$$

3. Aug 23, 2009

### nameVoid

can someone please confirm the method used to solve the second

4. Aug 23, 2009

### mathie.girl

If the second one is:

$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2}$$

Then you should be fine with the nth term test:

Last edited by a moderator: May 4, 2017
5. Aug 23, 2009

### zcd

If you're testing for conditional convergence, then the conditions are $$a_{n}$$ must be constantly decreasing (monotonic decreasing) and $$\lim_{n\to\infty}a_{n}=0$$. In the case of the second problem, the test you took would apply for both conditional and absolute convergence, so it's right.

6. Aug 24, 2009

### fmam3

For this one: $$\sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2}$$, you might want to be a little bit more careful about your argument; you've written a couple things down, and some are not entirely correct (but could be just the awkward formatting that's making things hard to read).

It seems to me you're confusing a few concepts here. You wrote $$\frac{n^2+3}{(2n-5)^2} \leq 1$$ and then wrote that $$\sum_{n=1}^{\infty}1 \to \infty$$ --- are you attempting to use the Comparison Test here? Because knowing that $$\frac{n^2+3}{(2n-5)^2} \leq 1$$ holds and even knowing that $$\sum_{n=1}^{\infty}1 \to \infty$$ says nothing about the convergence of $$\sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2}$$. Besides, the inequality you wrote down is wrong. Set $$n = 2$$ and you immediately have $$\frac{2^2 + 3}{(2\cdot2 - 5)^2} = \frac{4 + 3}{(-1)^2} = 7 \not\leq 1$$.

A good approach here would be to use the Alternating Series Test.

7. Aug 24, 2009

### g_edgar

Incorrect. Just for fun, can you do this limit correctly?

8. Aug 24, 2009

### nameVoid

$$\lim_{n->\infty}\frac{10}{(n!)^(1/n)} = 10$$
I assumed

9. Aug 24, 2009

### g_edgar

$n!$ is much bigger than $10^n$ when $n$ is large.

For example:

$$10000000000000000000000000000000000000000 = 10^{40}$$

$$815915283247897734345611269596115894272000000000 = 40! = 8.159 \times 10^{47}$$