Sequences, Series, Convergence and Divergence

[FaraDazed]

Homework Statement

Q1 Are the following sequences divergent or convergent as n tends to infinity.

a: \frac{5n+2}{n-1}

b: tan^{-1}(n)

c:\frac{2^n}{n!}

Q2 Evaluate:...

a: \sum_{n=1}^{\infty} 3^{\frac{n}{2}}

b: \sum_{n=1}^{99} (-1)^n

Q3 Find whether the following converge or diverge

a:\sum_{n=1}^{\infty} \frac{n-1}{n}

b:\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}

Homework Equations

|\frac{a_{n+1}}{a_n}|

The Attempt at a Solution

Most of these I have no clue on how to format it mathmatically correct but I have given it my best shot, Id be surprised if I have any correct mind you.

Q1a
<br /> \frac{5n+2}{n-1}=12,\frac{17}{2},\frac{22}{3},\frac{27}{4}...\\<br />
So it looks as tho it converges to 0 as n tends to infinity.

Q1b
<br /> tan^{-1}(n) = \frac{\pi}{4},1.107,1.25...<br />
From messing on the calculator I can see that it tends to pi/2 as n tends to infinity but don't know how to show it mathmatically.

Q1c
<br /> \frac{2^n}{n!}=2,2,1.33,0.66,0.266<br />
Again looks like it converges to 0 as n tends to infinity.

Q2a
<br /> \sum_{n=1}^{\infty} 3^{\frac{n}{2}}=1+\frac{1}{\sqrt{3}}+\frac{1}{3}+0.192+\frac{1}{9}<br />
This looks as though the limit is 2, as n tends to infinity

Q2b
<br /> \sum_{n=1}^{99} (-1)^n=1-1+1-1+1-1<br />
I noted that when n is even it is +1 and when it is odd it is -1 and since 99 is odd, then the limit is 0 as n tends to 99.

Q3a
\sum_{n=1}^{\infty} \frac{n-1}{n}=\frac{\frac{n}{n}-\frac{1}{n}}{\frac{n}{n}}=1-\frac{1}{n} \\<br /> =1-1+1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}...
here it seems it diverges to infinity as n tends to infinity (due to the +1's)

Q3b
<br /> \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}=1+0.353+0.192+...<br />
Here is looks like it converges to the limit of 2 as n tends to infinity.Sorry for the mass of questions, I am not sure about any of them so any advice would be much appreciated.

Thanks.

 

[member 428835]

hey! sorry, i didn't look through all the equations, but i'll offer some help on 1c) \frac{2^n}{n!}=\frac{2}{n}\cdot\frac{2}{(n-1)}\cdot\frac{2}{(n-2)}\cdot \cdot \cdot \cdot \frac{2}{3}\cdot\frac{2}{2}\cdot\frac{2}{1}. now what can we do (compare)...i'll let you think on this.

 

[Mark44]

FaraDazed,
Please limit the number of problems you post in a thread to one or, at most, two.

I have closed this thread. Feel free to start new threads with a problem or two in each.