Sequence and series of convergence problem: $2-2^{1/2}+2^{1/3}-2^{1/4}+2^{1/5}-2^{1/6}+\ldots$ We have a sequence/series problem. and i need help
Consider the following:
How can we show that $2-2^{1/2}+2^{1/3}-2^{1/4}+2^{1/5}-2^{1/6}+\ldots$ diverges?
I am so lost as to solving this problem. 
I understood the definition of convergence and divergence theorem.
any help would be appreciated. thank you
 A: *

*Because the terms $a_k=2^{1/k}$ do not tend to zero for $k \to \infty$ (instead they tend to $1$), the series cannot be convergent.
And simply: if a series is not "convergent" we call it "divergent" (by definition of the term "divergent"). - so: that's already how your question could be answered and the result be shown.      

*But one can proceed if desired: Since the signs oscillate we can possibly do some flattening/averaging of the partial sums.
The sequence of partial sums $s_k$ eventually oscillates between $  0.73306828$ and $1.7441314$, and if we allow/introduce the concept of divergent summation we can, for instance by Cesaro-summation, assign the divergent series the value $1.2386440...$ (which is simply the mean of the two oscillatory values), which shall then be valid in the same sense as we assign/accept values for the (similarly) divergent series of the Dirichlet $\eta$ (the "alternating $\zeta$") at zero or at negative arguments.
Since we are widely "used to" accept the value ${1 \over 1+x}$ for the alternating geometric series $1-x+x^2-x^3 \pm \cdots $ even for $x=1$ (which is already a divergent series!) and thus already basically think in terms of divergent summation, we can discuss the series $ s_a=(2^1-1)-(2^{1/2}-1)+(2^{1/3}-1) \pm \cdots $ and $s_b = 1-1+1-1 \pm \cdots ={1 \over 1+x}_{\mid x=1}= 1/2 $ (alternating geometric series )
and then evaluate the now convergent series $s_a = 0.73864396...$ (you must now show that this series is convergent!) and add $s=s_a + s_b=0.73864396...+1/2=1.23864396$ 
