Proving the Convergence of a Sequence: Monotonicity and Boundedness

[paniurelis]

Let's have a sequence $$x_n=\sum_{k=1}^{k=n}{\frac{1}{2^{\sqrt{k}}}$$.
We must prove it is convergent.
First thought, let's try to prove it is monotonic and bounded, which means convergence of sequence.
Monotonicity is easy, $$\forall n \in N: x_{n+1}-x_n = \frac{1}{2^{\sqrt{n+1}}} > 0$$
So, sequence is increasing. Next, I should prove it has an upper bound, but I am not able to come up with bigger sequence which would have a positive finite limit.
Any ideas ?

 

[morphism]

I think the best way to approach this problem is via the integral test. 2^(-sqrt(k)) is going very slowly to zero, so it's going to be tough to find anything that goes to zero slower than it but for which we can verify summability easily. (But it's 5:30am, so I'm probably missing something!)

 

[paniurelis]

Thanks morphism, it is a good idea. The problem is, I have an impression, the solution should need only the knowledge of limit theory and some elementary algebra, because I found this problem in the problem book for calculus I, before the chapters with problems about integrals...
IMHO, there is some clever trick I am not aware about...:-(
So, does anybody have more ideas ?

 

[VietDao29]

Thanks morphism, it is a good idea. The problem is, I have an impression, the solution should need only the knowledge of limit theory and some elementary algebra, because I found this problem in the problem book for calculus I, before the chapters with problems about integrals...
IMHO, there is some clever trick I am not aware about...:-(
So, does anybody have more ideas ?

So for any natural n large enough (say, n >= N), we have: 2ⁿ >= n⁴. You can prove this using the Limit: $$\lim_{x \rightarrow \infty} \frac{x ^ 4}{2 ^ x} = 0$$.

So, for any natual n > N, we have:

$$\sum_{k = 1} ^ {n} \frac{1}{2 ^ {\sqrt{k}}} = \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{k' = N} ^ {n} \frac{1}{2 ^ {\sqrt{k'}}} \leq \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{k' = N} ^ {n} \frac{1}{k' ^ 2}$$.

Can you go from here? :)

 

[paniurelis]

$$\exists N=16 \forall n > N : n^4 < 2^n$$
$$\forall n > N : \sum_{k = 1} ^ {n} \frac{1}{2 ^ {\sqrt{k}}} = \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{2 ^ {\sqrt{l}}} < \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{l ^ 2}<\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{(l-1)l}=\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \left(\frac{1}{l-1}-\frac{1}{l}\right)=$$
$$=\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \frac{1}{N}-\frac{1}{n}<\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \frac{1}{N}$$

So, increasing sequence has an upper bound, so it is convergent.
Q.E.D.

Is it correct ?
Thanks for help :-))) I am very happy now

 

[VietDao29]

$$\exists N=16 \forall n > N : n^4 < 2^n$$
$$\forall n > N : \sum_{k = 1} ^ {n} \frac{1}{2 ^ {\sqrt{k}}} = \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{2 ^ {\sqrt{l}}} < \sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{l ^ 2}<\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \frac{1}{l (l+1)}=\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \sum_{l = N+1} ^ {n} \left(\frac{1}{l}-\frac{1}{l+1}\right)=$$
$$=\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \frac{1}{N+1}-\frac{1}{n+1}<\sum_{k = 1} ^ {N} \frac{1}{2 ^ {\sqrt{k}}} + \frac{1}{N+1}$$

So, increasing sequence has an upper bound, so it is convergent.
Q.E.D.

Is it correct ?
Thanks for help :-))) I am very happy now

Yup, looks excellent. ^^! Well done.

 

[paniurelis]

I have corrected error in my post above, was $$\frac{1}{k^2}<\frac{1}{k(k+1)}$$,
corrected $$\frac{1}{k^2}<\frac{1}{(k-1)k}$$