- #1
FaraDazed
- 347
- 2
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.
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