Sum of a series, how do I find the sum? I'm stuck on the last few steps.

I'll just use S to denote sum from 1 to infinity.

Find the sum.
S[(1/(n^(1/3))) - (1/((n+1)^(1/3)))] (one over cube root n, minus 1 over cube root of (n plus 1))

First, use test for divergence:

lim [(1/n^(1/3)) - (1/(n+1)^(1/3))] = 0
x->infinity

So it might diverge or converge. Going further, by evaluating some terms:

S[(1/(n^(1/3))) - (1/((n+1)^(1/3)))] = ((1/1) - (1/2^(1/3))) + (1/(2^(1/3))) - (1/((3^(1/3))) + ...

I don't want to fill the whole page with a telescoping sum, butI from this, I have found out it's a telescoping sum. All terms after 1 cancel out. Thus, the series converges because everything after the 1 cancels (it converges to 1).

Now I don't know how to find the sum. I tried the property:

S(a_n + b_n) = S(a_n) + S(b_n), provided the two series converge. But upon testing for convergence of 1/(n^(1/3)), I found that it does not converge, so I can't use this. What do I do?

Integration doesn't seem like an option, but I'm not sure what to do. The sum looks like it is 1 because everything afterwards cancels, but I don't know if and how I can show this.

Can anybody please help?

Thank you.

You've done most of the hard work on this one.

Σ(n = 1 to ∞) [1/n^(1/3) - 1/(n+1)^(1/3)]
= lim(k→∞) Σ(n = 1 to k) [1/n^(1/3) - 1/(n+1)^(1/3)]
= lim(k→∞) [(1 - 1/2^(1/3)) + (1/2^(1/3) - 1/3^(1/3)) + ... + (1/k^(1/3) - 1/(k+1)^(1/3))]
= lim(k→∞) [1 - 1/(k+1)^(1/3)], since all other terms cancel in pairs
= 1 - 0
= 1 (convergent).

I hope this helps!

........ n
f(n) = ∑ [(1/(k^(1/3))) - (1/((k+1)^(1/3)))]
....... k=1

........ n ................... n+1
f(n) = ∑ [(1/(k^(1/3))) - ∑ (1/(k)^(1/3) ] = 1/(1)^(1/3) - 1/(n+1)^(1/3)
....... k=1 ................ k=2

Lim f(n) = 1
n→∞