Determine if the Series Converges and Explain approach Is the series Convergent? Explain and find Sum:-
$$\sum_{n=1}^{\infty} \ (2)^\frac{1}{n}-(2)^\frac{1}{n+1}$$
Thanks
 A: Show convergence
You can use the integral tests for convergence on this problem, because the integral of this function is fairly easy to determine. The integral test states that your series of the form:
$$ \sum_{n=1}^\infty f(n) $$
converges if and only if the solution to the following integral is finite:
$$ \int_1^\infty f(x) dx $$
In your case $f(x)=2^{\frac{1}{x}}-2^\frac{1}{x+1}$ yields the following answer for the integral:
$$ \log (2) \left(\text{li}(2)-\text{Ei}\left(\frac{\log (2)}{2}\right)\right)+2 \sqrt{2}-3$$
where $\text{li}$ is the logarithmic integral function and $\text{Ei}$ is the exponential integral.
which is finite ($\approx 0.624$ if you evaluate it) so your function converges.
Find sum
If you write out a part of the series you can see that almost all terms will cancel:
$$ \sum_{n=1}^{N} \ (2)^\frac{1}{n}-(2)^\frac{1}{n+1} = 2^\frac{1}{1}-2^\frac{1}{2}+2^\frac{1}{2}-2^\frac{1}{3}+2^\frac{1}{3}-2^\frac{1}{4}+....+2^\frac{1}{N}-2^\frac{1}{N+1}$$
All terms except the very last and the very first drop out so you get:
$$ \sum_{n=1}^{N} \ (2)^\frac{1}{n}-(2)^\frac{1}{n+1} = 2^\frac{1}{1}-2^\frac{1}{N+1}$$
now taking the limit for $N\to\infty$: $\lim_{N\to\infty} 2^{1/(N+1)} =1 $
so you get convergence to 1.
Obviously, the fact that you can find a finite value for this sum is already a proof in itself that the series is convergent.
A: Since
$$\sum_{n=1}^N 2^{1/n}-2^{1/(n+1)} = 2-2^{1/(N+1)}$$
and 
$$\lim_{N\to\infty} 2^{1/(N+1)} =1,$$
so this series is convergent and  converges to 1.
A: Hint: Consider a series $\sum\limits_nx_n$ where $x_n=a_n-a_{n+1}$ for every $n\geqslant1$, for some sequence $(a_n)_{n\geqslant1}$. 


*

*Express each partial sum $\sum\limits_{n=1}^Nx_n$ with $N\geqslant1$, as a simple function of $a_1$ and of another term of the sequence $(a_n)_{n\geqslant1}$. 

*Deduce a simple condition implying that $\sum\limits_{n=1}^Nx_n$ converges when $N\to\infty$.

*When this condition is met, express the limit $\sum\limits_{n=1}^\infty x_n$ in terms of  $(a_n)_{n\geqslant1}$.

