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 April 30th, 2011, 12:11 PM #1 Senior Member   Joined: Apr 2011 From: Recife, BR Posts: 352 Thanks: 0 Convergence of an infinite sum How can we determine the convergence of a sum when both the ratio and the root tests are inconclusive? Say we have the sum $\sum \limits_{n=2}^{\infty} \ \frac{(-1)^n}{\log(n)}$ Wolfram Alpha gives an "approximated sum", but from the plot I can't even tell if it's convergent.
 April 30th, 2011, 12:33 PM #2 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Convergence of an infinite sum proglote, This particular sum is called an 'alternating series' since the terms alternate in sign. For such series, write them as $S\=\ \sum_n (-1)^nb_n$ where bn ? 0. If $\text{ } 1) \quad \text{eventually } \forall n, \quad b_{n+1}\ \leq\ b_{n}, \text{ and}\\ \text{ } 2) \quad\lim_{n\rightarrow\infty} b_n\ =\ 0,$ then the series converges. Sometimes we can figure out the sum by clever tricks, but most of the time we cannot get a nice closed-form expression for the sum of an arbitrary series. -Ormkärr-
April 30th, 2011, 01:09 PM   #3
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Re: Convergence of an infinite sum

Quote:
 Originally Posted by Ormkärr proglote, This particular sum is called an 'alternating series' since the terms alternate in sign. For such series, write them as $S\=\ \sum_n (-1)^nb_n$ where bn ? 0. If $\text{ } 1) \quad \text{eventually } \forall n, \quad b_{n+1}\ \leq\ b_{n}, \text{ and}\\ \text{ } 2) \quad\lim_{n\rightarrow\infty} b_n\ =\ 0,$ then the series converges. Sometimes we can figure out the sum by clever tricks, but most of the time we cannot get a nice closed-form expression for the sum of an arbitrary series. -Ormkärr-
These two conditions are met, but do they ascertain that the series converge to a determinate number? Can't the sum be undefined in an interval, the same way as -1 + 1 - 1 + 1..?

 April 30th, 2011, 01:19 PM #4 Senior Member   Joined: Jun 2010 Posts: 618 Thanks: 0 Re: Convergence of an infinite sum proglote, Convergence is defined as meaning that the sum gets arbitrarily close to some definite number (its sum) as more and more terms are added. So a series like $S\=\ 1\ -\ 1\ +\ 1\ -\ 1\ +\ \cdots$ does not converge, since the partial sums are 0 and 1, depending on whether an even or odd number of terms are added, and these values certainly do not form a convergent sequence. -Ormkärr-
June 20th, 2011, 12:26 AM   #5
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Re: Convergence of an infinite sum

Quote:
 Originally Posted by proglote How can we determine the convergence of a sum when both the ratio and the root tests are inconclusive?

Criteria are for example :

*) $\sum_{n=1}^\infty a_n$ converges if $\lim_{n-=>\infty} n\cdot a_n=0$ (?)

*) if $|a_n|=<=|b_n|$ and $\{b_n\}$ converges, then
the series in a converges too, if it's bigger than a divergent series's terms, it diverges.

*) $\sum_{n=1}^\infty a_n$ and $\int_c^\infty a(n)dn$ have the same convergence.

 July 3rd, 2011, 11:29 PM #6 Member   Joined: Jun 2010 Posts: 80 Thanks: 0 Re: Convergence of an infinite sum For the first criterion I'm not sure about, based on the following proof, but may contain some pitfall : The series converges if $\forall\delta,\exists N(\delta)|\textrm{if}\quad n>N(\delta) then \mid S_{n+m}-S_n\mid<\delta$ since $m*a_{m+N(\delta)-p}\le\mid S_{n+m}-S_n\mid<\delta$, if p=0 then it induces an hypothesis about the fact that the terms are decreasing, else $p, p is to determine the minimum of the terms between m and m+N, if it exists (proof?) taking the limit as $\delta->0$, we find that $n*a_n$ tends towards 0 since N is independent of m.
 July 4th, 2011, 02:11 AM #7 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Convergence of an infinite sum Unfortunately your criteria don't work. A counterexample for the first is $a_n=\frac1{n\log n}.$ You need $b_n$ to converge absolutely for the second criterion. As it stands, it doesn't work since convergence does not imply absolute convergence (counterexample: $a_n=\frac1{2n},\ b_n=\frac{(-1)^n}n.$ For the third you need a lot more conditions on $a(n)$ than just $a(n)=a_n$ for integer n - otherwise we could easily have something like $a(n)=1-\cos(2\pi n),$ where $a(n)=0$ for integer n but the integral clearly does not converge. Conversely, if $a(n)=\frac{\cos(2\pi n)}n$ then $\int_1^\infty a(n)\,dn$ is finite but $\sum_{n=1}^\infty a(n)$ diverges.
 July 4th, 2011, 12:12 PM #8 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Convergence of an infinite sum The series converges to around .9243 Can this be shown?.
 July 9th, 2011, 12:33 AM #9 Member   Joined: Jun 2010 Posts: 80 Thanks: 0 Re: Convergence of an infinite sum I don't know which series you mean. The series 1/(n*log(n)) adds up to 3.8 at 2 billions, however it's very slow.
 July 9th, 2011, 12:51 AM #10 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Convergence of an infinite sum $\sum_{n=2}^\infty\frac{1}{n\log n}$ diverges - I don't know what all this 3.8 nonsense is about $\sum_{n=2}^N\frac1{n\log n}=\int_2^{N+1} \frac1{\lfloor n\rfloor\log\lfloor n\rfloor}\,dn\,>\,\int_2^{N+1}\frac1{n\log n}\,dn=\log(\log(N+1))-\log(\log(2))$ This is a case of where you can use the corresponding integral to check convergence - this is because the integrand is monotonic.

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