Series Convergence: Alternating and Absolute Convergence Explained

[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

 

[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

 

[nameVoid]

can someone please confirm the method used to solve the second

 

[mathie.girl]

If the second one is:

<br /> \sum_{n=1}^{\infty}(-1)^n\frac{n^2+3}{(2n-5)^2} <br />

Then you should be fine with the n^th term test:

If[/PLAIN] \lim_{n \to \infty} a_n \neq 0 or if the limit does not exist, then \sum_{n=1}^\infty a_n diverges.

 

[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.

 

[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.

 

[g_edgar]

\lim_{n-&gt;\infty}\sqrt[n]{\frac{10^n}{n!}}&gt;1

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

 

[nameVoid]

\lim_{n-&gt;\infty}\frac{10}{(n!)^(1/n)} = 10
I assumed

 

[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}