Series and Sequence Help: Convergence and Limits with r = 11/22

[JessicaSTAR]

hi everyone, I am having trouble with a few problems, i was wondering if anyone can help me thank.
1.) given r= 11/22

a.)consider the sequence {nr^r}. if convergent find limit. if divergent find if it goes to inf or minus inf. or div otherwise
lim nr^r = ?
n->infinity

b.)take my word for it that it can be shown that
sigma i=1 to n ir^i = (n(r^(n+2))-(n+1)(r^(n+1))+r)/((1-r)^2)

now consider the series sigma n=1 to infinity nr^r
sigma n=1 to infinity nr^r = ? 2.) An= 8n/(6n+13)
the series sigma n=1 to infinity (An) =?
if convergent, find sum. if divergent find it it goes to positive or negitive.and 3.) An= 50/(5^n)
find whether {An} is convergent, if so, find limit.

any help would be great. thanks.

 

[siddharth]

hi everyone, I am having trouble with a few problems, i was wondering if anyone can help me thank.
1.) given r= 11/22

You mean r=0.5?

a.)consider the sequence {nr^r}. if convergent find limit. if divergent find if it goes to inf or minus inf. or div otherwise
lim nr^r = ?
n->infinity

If r=0.5, then what's \lim_{n\rightarrow \infty} n \sqrt{0.5}

b.)take my word for it that it can be shown that
sigma i=1 to n ir^i = (n(r^(n+2))-(n+1)(r^(n+1))+r)/((1-r)^2)
now consider the series sigma n=1 to infinity nr^r
sigma n=1 to infinity nr^r = ?

You have lost me here. If r=0.5, how is the series convergent in the first place?

2.) An= 8n/(6n+13)
the series sigma n=1 to infinity (An) =?
if convergent, find sum. if divergent find it it goes to positive or negitive.

What does the divergence test tell you?

and 3.) An= 50/(5^n)
find whether {An} is convergent, if so, find limit.

You have already stated how to find if a sequence is convergent or divergent. Why not apply it here?

 

[JessicaSTAR]

sorry for the first one a.) and b.) instead or nr^r i meant nr^n

ok for 2.) i got infinity

and 3.) would go to minus infinity so would it be divergent?

 

[JessicaSTAR]

i still need help on 1a and b and 3, anyhelp would be good, thanks.