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## Homework Statement

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

a: [itex]\frac{5n+2}{n-1}[/itex]

b: [itex]tan^{-1}(n)[/itex]

c:[itex]\frac{2^n}{n!}[/itex]

Q2 Evaluate:....

a: [itex]\sum_{n=1}^{\infty} 3^{\frac{n}{2}} [/itex]

b: [itex]\sum_{n=1}^{99} (-1)^n [/itex]

Q3 Find whether the following converge or diverge

a:[itex]\sum_{n=1}^{\infty} \frac{n-1}{n}[/itex]

b:[itex]\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}[/itex]

## 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

[tex]

\frac{5n+2}{n-1}=12,\frac{17}{2},\frac{22}{3},\frac{27}{4}...\\

[/tex]

So it looks as tho it converges to 0 as n tends to infinity.

Q1b

[tex]

tan^{-1}(n) = \frac{\pi}{4},1.107,1.25...

[/tex]

From messing on the calculator I can see that it tends to pi/2 as n tends to infinity but dont know how to show it mathmatically.

Q1c

[tex]

\frac{2^n}{n!}=2,2,1.33,0.66,0.266

[/tex]

Again looks like it converges to 0 as n tends to infinity.

Q2a

[tex]

\sum_{n=1}^{\infty} 3^{\frac{n}{2}}=1+\frac{1}{\sqrt{3}}+\frac{1}{3}+0.192+\frac{1}{9}

[/tex]

This looks as though the limit is 2, as n tends to infinity

Q2b

[tex]

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

[/tex]

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

[tex]\sum_{n=1}^{\infty} \frac{n-1}{n}=\frac{\frac{n}{n}-\frac{1}{n}}{\frac{n}{n}}=1-\frac{1}{n} \\

=1-1+1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}...[/tex]

here it seems it diverges to infinity as n tends to infinity (due to the +1's)

Q3b

[tex]

\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}=1+0.353+0.192+.....

[/tex]

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.

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