convergence of sum $\sum_{j=1}^\infty\frac{1}{2}2^{-j/2}\sqrt{j+1}$

Question

I need to prove that for every $\epsilon>0$ there is $J$ large enough such that $$\sum_{j=J}^\infty\frac{1}{2}2^{-j/2}\sqrt{j+1}<\epsilon$$ but I can't seem to find out why since the set seems to be greater than the set of reciprocals of 2, which is convergent. Anyone has a hint?

Answer 1

Let's prove by induction that $j+1 \le 2^{\frac j 2}$ for $j \ge 6$. If $j=6$ then indeed $7 \le 8$. The induction step:

$$j+2 = (j+1) + 1 \le 2^{\frac j 2} + 1 \le 2^{\frac j 2} + 2^{\frac 6 2} (\sqrt 2 -1) = 2^{\frac j 2} + 2^{\frac j 2} (\sqrt 2 -1) = 2^{\frac {j+1} 2} .$$

This means that for $J \ge 6$ we may write

$$\sum _{j=J} ^\infty \frac 1 2 2^{-\frac j 2} \sqrt {j+1} \le \sum _{j=J} ^\infty \frac 1 2 2^{-\frac j 2} \sqrt {2^{\frac j 2}} = \frac 1 2 \sum _{j=J} ^\infty \frac 1 {2 ^{\frac j 2 - \frac j 4}} = \frac 1 2 \sum _{j=J} ^\infty \frac 1 {\sqrt[4] 2 ^j} = \frac 1 {\sqrt[4] 2 ^J} \frac 1 {2 - \frac 2 {\sqrt[4] 2}} .$$

Now, a sufficient condition for what you want is to impose that

$$\frac 1 {\sqrt[4] 2 ^J} \frac 1 {2 - \frac 2 {\sqrt[4] 2}} < \epsilon ,$$

which means that

$$J > \log _{\sqrt[4] 2} \frac 1 {\left( 2 - \frac 2 {\sqrt[4] 2} \right) \epsilon} = - \log _{\sqrt[4] 2} \left( \left( 2 - \frac 2 {\sqrt[4] 2} \right) \epsilon \right) .$$

Answer 2

It follows from that the series itself being convergent. Assume you have a convergent series $A = \sum a_k$ then you will have by definition that $\lim_{N\to\infty}\sum_0^{N-1} a_k = A$ which means again by definition that $|A-\sum_0^{N-1}a_k|=|\sum_N^\infty a_k|<\epsilon$ for large enough $N$.

So what you need to show is that $\sum {1\over2}2^{-j/2}\sqrt{j+1}$ is convergent, but that can be shown using the ratio criterion. The ratio between successive terms is ${1\over\sqrt2}\sqrt{j+2\over j+1} \to {1\over\sqrt2}$ as $j\to\infty$.