Krdytkyu

So i have to show the value of the series

$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i$$

For this to exist it has to be convergent, so i checked if it is divergent using the limit test:

$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=1$$

Which clearly shows, that it is divergent. Now to the problem: Most/all of my fellow students claim to have a seriesvalue. Is the limit test not conclusive evidence for a series to be divergent?

Edit: I am sorry, I messed up. It should be a sum from n to infinity and not from i to n

I calculated the value of the series: $S=sumlimits_{k=0}^infty left(1-(-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$

Form $S$ in the following way:

$S=sumlimits_{k=0}^infty left(frac 13right)^k-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k=frac {3}{2}-sumlimits_{k=0}^infty left((-1)^{frac{k(k+1)}2}right)cdot left(frac 13right)^k$

Separate the sum accordance with even and odd $k$:

If k even then $k=2m$, if odd then $k=2m+1$

Substitute them into S we get:

$S=frac {3}{2}-big(sumlimits_{k=0}^infty (-1)^{m}cdot left(frac 13right)^{2m}+sumlimits_{k=0}^infty (-1)^{3m+1}cdot left(frac 13right)^{2m+1}big)$

Repeat the separation for m values:

If m even then $k=2r$, if odd then $k=2r+1$

we have:

$S=frac {3}{2}-sumlimits_{r=0}^infty big[frac{(-1)^{2r}}{3^{4r}}+frac{(-1)^{2r+1}}{3^{4r+2}}+frac{(-1)^{6r+1}}{3^{4r+1}}frac{(-1)^{6r+4}}{3^{4r+3}}big]=frac {3}{2}-sumlimits_{r=0}^infty big[frac{1}{3^{4r}}big(1-frac{1}{9}-frac{1}{3}+frac{1}{27}big)big]$

Summing the geometrical series and perform the simplifications we get that:

$S=frac{9}{10}$

Using this method you can easily determine the value sum from k=0 to n.

We are dealing with a sum not with a series, it is a series if we are dealing with its limit as $nto infty$.

To evaluate the sum note that

• $(-1)^{frac{i(i+1)}2}=1,-1,-1,1,1,-1,-1,1,ldots$

then for $n=4k$

$$sum_{i=0}^n left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^i=sum_{i=0}^{k-1} left(frac 13right)^{1+4k}+sum_{i=0}^{k-1} left(frac 13right)^{2+4k}$$

With reference to the series note that

$$lim_{n to infty} left(1-(-1)^{frac{n(n+1)}2}right)cdot left(frac 13right)^n=0$$

therefore we can't conclude form here that the series diverges.

Otherwise the series clearly converges since

$$sum_{i=0}^infty left(1-(-1)^{frac{i(i+1)}2}right)cdot left(frac 13right)^ile sum_{i=0}^{infty} left(frac 13right)^{i}$$

and the RHS is a convergent geometric series.