Question

is the series absolutely convergent, convergent, or divergent? (alternating series )

Answer 1

\[\sum_{n=1}^{\infty} -1^n n ^{-1}\ln(n+3) \]
\[\sum_{n=1}^{\infty} \frac{ (n+2)^{n} }{ 2^{n}^{2} } \]

Answer 2

\[\sum_{n=1}^{\infty} \frac{ (n+2)^{n} }{ 2^{n^2}} \]

Answer 3

\(\color{#000000}{ \displaystyle \sum_{ n=1 }^{ \infty } (-1)^n\left[\color{blue}{\frac{\ln(n+3)}{n}}\right]}\)

Answer 4

As I have said before, if there is a term n=j from which the series converge (like n=100, or n=98264 or any n=k), then this series will as well converge starting from n=1.

(provided that none of the terms are singular)

Answer 5

If you start from n=1000 for example, then you have a monotonically decreasing sequence \(A_n\), and the series alternates.

Answer 6

So starting from n=1000 the series would converge, and therefore .....

Answer 7

But, the question is, would this series converge without alternation?

Answer 8

if it does without alternation, then it converges absolutely.
if it does NOT converge without alternation, then it converges conditionally.

Answer 9

Will \(\color{#000000}{ \displaystyle \sum_{ n=1 }^{ \infty } \left[\color{blue}{\frac{\ln(n+3)}{n}}\right]}\) converge ?

Answer 10

As n tends to infinity you know that:
\(\color{#000000}{ \displaystyle \color{blue}{\frac{(n+2)^n}{2^{n^2}}}>\color{blue}{\frac{(n+n)^n}{2^{n^2}}} }\)

\(\color{#000000}{ \displaystyle \color{blue}{\frac{(n+n)^n}{2^{n^2}}} =\color{blue}{\frac{(2n)^n}{2^{n^2}}} =\color{blue}{\frac{2^nn^n}{2^{n^2}}}=\color{blue}{\frac{n^n}{2^{n^2}2^{-n}}}=\color{blue}{\frac{n^n}{2^{(n-1)n}}} }\)

\(\color{#000000}{ \displaystyle =\color{blue}{\frac{n^n}{\left(2^{n-1}\right)^n}} }\)

Answer 11

Oh my sign is wrong

Answer 12

the first equation should say <.
a simple typo, sorry

Answer 13

So....

Would you agree that
\(\color{#000000}{ \displaystyle \sum_{n=j}^\infty \left(\frac{n}{2^{n-1}}\right)=\rm converges }\) ?

Now, since
\(\color{#000000}{ \displaystyle \left(\frac{n}{2^{n-1}}\right)<1 }\)

therefore
\(\color{#000000}{ \displaystyle \left(\frac{n}{2^{n-1}}\right)>\left(\frac{n}{2^{n-1}}\right)^n }\)
(Because raising number x to larger powers if |x|<1 makes the result smaller)

and so, all the more so
\(\color{#000000}{ \displaystyle \sum_{n=j}^\infty \left(\frac{n}{2^{n-1}}\right)^n=\rm converges }\)
(because this series is smaller than the previous)

and since (as I deduced)
\(\color{#000000}{ \displaystyle \frac{(n+2)^n}{2^{n^2}}>\left(\frac{n}{2^{n-1}}\right)^n }\)

therefore
Would you agree that

\(\color{#000000}{ \displaystyle \sum_{n=j}^\infty \left(\frac{n}{2^{n-1}}\right)=\rm converges }\)

Since
\(\color{#000000}{ \displaystyle \left(\frac{n}{2^{n-1}}\right)<1 }\)

therefore
\(\color{#000000}{ \displaystyle \left(\frac{n}{2^{n-1}}\right)>\left(\frac{n}{2^{n-1}}\right)^n }\)

and so, all the more so
\(\color{#000000}{ \displaystyle \sum_{n=j}^\infty \left(\frac{n}{2^{n-1}}\right)^n=\rm converges }\)

and since (as I deduced)
\(\color{#000000}{ \displaystyle \frac{(n+2)^n}{2^{n^2}}>\left(\frac{n}{2^{n-1}}\right)^n }\)

therefore
\(\color{#000000}{ \displaystyle \sum_{n=j}^\infty \frac{(n+2)^n}{2^{n^2}}=\rm converges }\) ! ■

Answer 14

and the first series converges, like any polynomial over exponential.

In particular, you can start from n=1000000, and compare the polynomial numerator to 1.5^n (because at that starting point,and on forever, 1.5^n is greater than your polynomial), and thus you get (1.5^n/2^n)=(1.5/2)^n

so your series is behaving smaller than a geometric series with |r|<1.

Answer 15

I said something twice there.... hope it's not too confusing. good luck.

Answer 16

i still can't figure out what the series is

Answer 17

is it this:
\[\sum_{n=1}^{\infty} \frac{ (n+2)^{n} }{ 2^{n^2}}\]?

Answer 18

yes.

Answer 19

and the first one as well.

Answer 20

\(\displaystyle \sqrt[n]{\frac{ (n+2)^{n} }{ 2^{n^2}}}=\frac{n+2}{2^n}\to 0<1\) as \(n\to\infty\)

Answer 21

oh, I was overthought that:)
N-th root test:)