do these series converge absolutely, conditonally, or diverge: cosn/n^2;...
- Thread starter Spencer
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CogitoErgoCogitoSum
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- Jul 19, 2008
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Assuming the index value starts at n=0, or at n=1 if mathematically necessary....
the second one is particularly easy
(-1)^n 3^n/n! =
(-3)^n/n! =
e^-3 .......... via Maclaurin series expansion of e^x.
So the second series is convergent. If you ignored the alternating sign you would have had ? 3^n / n! = e^3. So its absolutely convergent.
If you didnt notice that, or didnt know that, then Im sure this series passes the ratio test. So try that.
====
? cos(n?)/n
This is relatively easy as well. Since n must be an integer value, cos(n?) must take a value of +1 or -1. This is an alternating series in disguise. It is exactly equal to:
? (-1)^n/n
By the alternating series test this is convergent. Of course, absolutely this is just a harmonic and is divergent.
If youre interested, this series equals ln(½)
======
? cos
/n²
Similar reasoning as before. The value of cos is always stuck (pinned, squeeze theorem) between +1 and -1. This series will possess negative terms and terms whose numerator is smaller in magnitude than its maximum of one... so the series will be smaller than if you took the numerator as +1 for all terms. If you look at the absolute series 1/n², you see this is a p-series with p=2. This is absolutely convergent.
The sum of a p-series with p=2 is going to be ?²/6... that means that your series in question will have a sum S that is:
-?²/6 < S < ?²/6
Im not entirely sure how it was done, but wolframalpha.com evaluates this series to exactly:
1/4 + ?²/6 - ?/2
=====
(-1)^n/rt(n+1)
This is an alternating series. Ignore the 'rt', whatever that is... constants maybe? Factor that out. The term (n+1) is a constantly growing value in the denominator. The series passes the alternating series test. Absolute series is nothing more than a harmonic horizontally shifted over by one.
the second one is particularly easy
(-1)^n 3^n/n! =
(-3)^n/n! =
e^-3 .......... via Maclaurin series expansion of e^x.
So the second series is convergent. If you ignored the alternating sign you would have had ? 3^n / n! = e^3. So its absolutely convergent.
If you didnt notice that, or didnt know that, then Im sure this series passes the ratio test. So try that.
====
? cos(n?)/n
This is relatively easy as well. Since n must be an integer value, cos(n?) must take a value of +1 or -1. This is an alternating series in disguise. It is exactly equal to:
? (-1)^n/n
By the alternating series test this is convergent. Of course, absolutely this is just a harmonic and is divergent.
If youre interested, this series equals ln(½)
======
? cos
/n²Similar reasoning as before. The value of cos
is always stuck (pinned, squeeze theorem) between +1 and -1. This series will possess negative terms and terms whose numerator is smaller in magnitude than its maximum of one... so the series will be smaller than if you took the numerator as +1 for all terms. If you look at the absolute series 1/n², you see this is a p-series with p=2. This is absolutely convergent.The sum of a p-series with p=2 is going to be ?²/6... that means that your series in question will have a sum S that is:
-?²/6 < S < ?²/6
Im not entirely sure how it was done, but wolframalpha.com evaluates this series to exactly:
1/4 + ?²/6 - ?/2
=====
(-1)^n/rt(n+1)
This is an alternating series. Ignore the 'rt', whatever that is... constants maybe? Factor that out. The term (n+1) is a constantly growing value in the denominator. The series passes the alternating series test. Absolute series is nothing more than a harmonic horizontally shifted over by one.