Question

Can you help me find the sum of series ((-1)^n) / (3n + 1) ?

Answer 1

\[\sum_{n=0}^{Infinity}\frac{ (-1)^n }{ 3n +1 }\]

Answer 2

have you tried writing out the first few partial sums to see if you can spot a pattern?

Answer 3

Yes, I have, and I get a sequence \[1-\frac{ 1 }{ 4 } +\frac{ 1 }{ 7}-\frac{ 1 }{ 10 }+...+\frac{ (-1)^n }{ 3n+1 }\] which limit I can't figure out.

Answer 4

the limit is not an elementary outcome ....

what level of math is this?

Answer 5

First year college.

Answer 6

which course?

Answer 7

Analysis 1, pure math

Answer 8

hmmm

the sum results in a transcendental number:\[\frac19(\sqrt{3}\pi+\log(8))\]

so the best you could do is prolly approximating the limit. What are some methods at your disposal?

Answer 9

I got slightly different solution using Wolfram Mathematica (there's also -9 in the brackets). Anyhow, I can use differentiation (though I'm not sure it'd help), MacLaurin polynomial/series and other basic methods.

Answer 10

i believe differentation plays into the Maclaurin method .... finding successive derivatives to develop a polynomial representation.

a stray thought i had was to use cos(n pi) instead of (-1)^n, find a relatively simply partial sum like S5 and average out the rest with integrations from 5 to inf, and 6 to inf

Answer 11

S5 = 0.80728021978...

(0.000829829-0.00116357)/2 = -0.0001668705

0.80728021978 -0.0001668705 = 0.80711334928
0.83564884826

but at S5, that doesnt seem to approximate it out too well

Answer 12

This sum was needed to be calculated on the written part of the exam from a few years ago where we can't use calculators and such, so I think that there's a way to rewrite that fraction somehow in order for the sum to be easier to find.

Answer 13

if we ignore the -1 and just write up a division


\(1-3n+9n^2-27n^3+...\)
------------------
1+3n ) 1
-(1+3n)
--------
-3n
-(-3n-9n^2)
----------
9n^2
.....

but i cant see where that would be useful at the moment

Answer 14

for this test, did you need an exact result?

Answer 15

yes, unfortunately

Answer 16

without ever have taken the course, im wondering if there is a relatively simple way to split out the positive and negative portions. but that end result is still daunting

Answer 17

@oldrin.bataku could you look into this series, please?

Answer 18

@amistre64 thank you for the effort :)

Answer 19

thnx for the challenge :)

Answer 20

some equivalent setups are
\[\sum_0 \frac{1}{6n+1}-\frac{1}{6n+4}\]
\[\sum_0 \frac{1}{6n+1}-\frac12\sum_0 \frac{1}{3n+2}\]

Answer 21

wonder if splitting in to even and odd helps? probably not, just wondering

Answer 22

then partial fractions or something?

Answer 23

the final results are so obfuscated to me that a clear path is obscured to me

Answer 24

yeah you got a \(\pi\) and log and a radical

Answer 25

sounds like a party

Answer 26

I've solved this :) The point was to use series \[\sum_{n=0}^{Infinity} \frac {(-1)^n} {3n+1} x ^{3n+1}\] and differentiate it to get \[\sum_{n=0}^{Infinity} (-1)^n x^{3n}\] which is MacLaurin's series for \[\frac{1} {1+x^3}\] Now find integral \[\int\limits_{0}^{x} \frac {1} {1+t^3}dt\] and it's going to be some function of x, and when we let \[\lim_{x \rightarrow 1}\] of that function it's going to give us the solution \[\frac{ 1 }{ 3 }\ln 2 + \frac {\pi} {3\sqrt3}\] P.S. I just had to share this with you :)

Answer 27

interesting technique fer sure

Answer 28

Right you should be able to recognize each term as a mere derivative. @amistre64 good idea but unfortunately the partial fractions will not telescope at all...