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frixis
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Series and Sequence Convergence... final on monday please help!
determine whether the sequence converges or diverges and find its limits
1(1+\frac{1}{n})^n
2 cos\pin
the overbar indicates that the digits underneat repeat indefinitely. Express as a series and find the rational number(since i can't find the latex for overbar, the brackets represent the overbar)
0.(23)
3.2(394)
show divergence or convergence by the nth term text
\displaystyle\sum_{n=2}^{\infty} \frac{1}{n\sqrt{{n^2}-1}}
use a basic comparision test to determine convergence
\displaystyle\sum_{n=1}^{\infty} \frac{\arctan n}{n}
determine convergence
1\displaystyle\sum_{n=1}^{\infty} \frac{2n+n^2}{n^3+1}
2\displaystyle\sum_{n=1}^{\infty} \frac{n^2+2^n}{n+3^n}
3\displaystyle\sum_{n=1}^{\infty} \frac{\mathrm{ln} n}{n^3}
4 \displaystyle\sum_{n=2}^{\infty} {\frac{1}{{n^3}{\sqrt[3]{\mathrm{ln}n}}}
5\displaystyle\sum_{n=1}^{\infty} n{\tan{\frac{1}{n}}
6\displaystyle\sum_{n=1}^{\infty} {(1+\frac{1}{n})}^n
7 1+\frac{1\cdot 3}{2!}+\frac{1\cdot 3\cdot 5}{3!}+...+\frac{1.3.5...(2n-1)}{n!}+...
find every real no k for which the series converges
\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^k{\mathrm{ln}n}}
use the proof of the integral test to show that, for every positive int n >1
{\mathrm{ln}(n+1)}{<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}<1+{\mathrm{ln}n}
determine wheter the series is absolotely convergent, conditionally convergent or divergent.
\displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{\cospin}{n}
show that the alternating series converges for ever positive integer k
\displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{{(\mathrm{ln}n)}^k}{n}
find the interval of convergence of the power series
1 \displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{x^n}{\sqrt{n}} can do the whole getting to the interval just can't decide whether or not it converges or diverges at -1 and 1 which are the intervals endpoints
2 \displaystyle\sum_{n=0}^{\infty} {\frac{n!}{100^n} x^n} issues with endpoints again
3 \displaystyle\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(-4)^n}
find the radius of convergence of the power series for positive integers c and d
\displaystyle\sum_{n=0}^{\infty} {\frac{{(n+c)}!}{n!{(n+d)}!} x^n
if the interval of convergence of \displaystyle\sum {a_n}{x^n} is (-r,r], prove that the series is conditionally convergent at r
um.. okay so i have to put down the attempts to all the questions? its a culmination of all issues...
for the decimals ones... i did the whole sum = a/1-r thing... for some reason i was doing something wrong...
for all convergence ones i always tried the ratio test or the root test... but i still had issues with the convergence... specially when ln comes in... i don't know y...
and the last one which is the most recent... i deduced that the ratio of the n+1 term and the n term is 1/r but then i tried putting that in the equation and i was pretty lost... because u'd get 1 and... yeah... okay...
so if i need to post all my attempts please let me know...
Homework Statement
determine whether the sequence converges or diverges and find its limits
1(1+\frac{1}{n})^n
2 cos\pin
the overbar indicates that the digits underneat repeat indefinitely. Express as a series and find the rational number(since i can't find the latex for overbar, the brackets represent the overbar)
0.(23)
3.2(394)
show divergence or convergence by the nth term text
\displaystyle\sum_{n=2}^{\infty} \frac{1}{n\sqrt{{n^2}-1}}
use a basic comparision test to determine convergence
\displaystyle\sum_{n=1}^{\infty} \frac{\arctan n}{n}
determine convergence
1\displaystyle\sum_{n=1}^{\infty} \frac{2n+n^2}{n^3+1}
2\displaystyle\sum_{n=1}^{\infty} \frac{n^2+2^n}{n+3^n}
3\displaystyle\sum_{n=1}^{\infty} \frac{\mathrm{ln} n}{n^3}
4 \displaystyle\sum_{n=2}^{\infty} {\frac{1}{{n^3}{\sqrt[3]{\mathrm{ln}n}}}
5\displaystyle\sum_{n=1}^{\infty} n{\tan{\frac{1}{n}}
6\displaystyle\sum_{n=1}^{\infty} {(1+\frac{1}{n})}^n
7 1+\frac{1\cdot 3}{2!}+\frac{1\cdot 3\cdot 5}{3!}+...+\frac{1.3.5...(2n-1)}{n!}+...
find every real no k for which the series converges
\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^k{\mathrm{ln}n}}
use the proof of the integral test to show that, for every positive int n >1
{\mathrm{ln}(n+1)}{<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}<1+{\mathrm{ln}n}
determine wheter the series is absolotely convergent, conditionally convergent or divergent.
\displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{\cospin}{n}
show that the alternating series converges for ever positive integer k
\displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{{(\mathrm{ln}n)}^k}{n}
find the interval of convergence of the power series
1 \displaystyle\sum_{n=1}^{\infty} {(-1)}^n \frac{x^n}{\sqrt{n}} can do the whole getting to the interval just can't decide whether or not it converges or diverges at -1 and 1 which are the intervals endpoints
2 \displaystyle\sum_{n=0}^{\infty} {\frac{n!}{100^n} x^n} issues with endpoints again
3 \displaystyle\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(-4)^n}
find the radius of convergence of the power series for positive integers c and d
\displaystyle\sum_{n=0}^{\infty} {\frac{{(n+c)}!}{n!{(n+d)}!} x^n
if the interval of convergence of \displaystyle\sum {a_n}{x^n} is (-r,r], prove that the series is conditionally convergent at r
The Attempt at a Solution
um.. okay so i have to put down the attempts to all the questions? its a culmination of all issues...
for the decimals ones... i did the whole sum = a/1-r thing... for some reason i was doing something wrong...
for all convergence ones i always tried the ratio test or the root test... but i still had issues with the convergence... specially when ln comes in... i don't know y...
and the last one which is the most recent... i deduced that the ratio of the n+1 term and the n term is 1/r but then i tried putting that in the equation and i was pretty lost... because u'd get 1 and... yeah... okay...
so if i need to post all my attempts please let me know...
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