Simplifying $\frac{a}{2^a}$ - Limit to 3/4?

[nobahar]

Is it possible to simplify this?

$$\frac{a}{2^a}$$

It's actually part of a limit:

$$2\lim_{n\rightarrow\infty}(\sum_{a=4}^{n}\frac{a}{2^a})$$

The limit I'm hoping tends to 3/4 as n tends to infinity. It certainly appears to do so when I plotted a graph of a/(2^a).

Any little hints? Tried logs but got nowhere: I still end up with an 'a' on the top and bottom. I'm assuming that I need to elliminate it from either the numerator or the denominator.
Thanks in advance.

 

[nobahar]

Okay, I got:
$$10^{\log{a}-a.C}$$

Where $$C=log{2}$$.
I don't know what to do with this.

 

[JG89]

Does this help:

$$\frac{d}{da} \left ( \frac{-1}{2^a} \left ) = \frac{a}{2^{a + 1}}$$

 

[nobahar]

I couldn't get the same result. I got:
$$\frac{d}{da} \left (\frac{-1}{2^a} \left ) = \frac {d}{da} (-1 * 2^{-a}) = \frac{\ln {2}}{2^a}$$
Because:
$$f(a)= \frac{-1}{2^a} = -1 * 2^{-a}$$

$$\ln {f(a)} = -1 * -a * \ln {2}$$

$$\frac{d}{da} \left (\ln {f(a)} \left ) = \ln {2} * \frac{d}{da} (a)$$

$$\frac {1}{f(a)} * \frac{d}{da} f(a) = 1 * \ln {2}$$

$$\frac{d}{da} f(a) = f(a) * \ln {2} = \frac{-1}{2^a} * \ln {2} = \frac{-\ln {2}}{2^a}$$

Is my calculus out?

Am I trying to establish the following?:
$$\int_4^\infty (\frac{a}{2a}) da$$

Any help much appreciated, thanks so much for the responses.

 

[Borek]

I couldn't get the same result. I got:
$$\frac{d}{da} \left (\frac{-1}{2^a} \left ) = \frac {d}{da} (-1 * 2^{-a}) = \frac{\ln {2}}{2^a}$$

Wolfram alpha says you are right:

http://www.wolframalpha.com/input/?i=d(-1/2^a)/da

 

[Mark44]

You could approximate
$$2 \sum_{a=4}^{\infty}\frac{a}{ 2^a}$$
by this integral:
$$2 \int_4^{\infty} \frac{x dx}{2^x}$$
assuming that both converge.

 

[nobahar]

Thanks for the replies guys!
I'm really sorry, I don't understand.
I thought
$$\sum_4^\infty \left (\frac{a}{2^a} \left )$$ would be a less accurate approximation of $$\int_4^\infty \left (\frac{a}{2^a} \left ) da$$;
and that
$$\int_4^\infty \left (\frac{a}{2^a} \left ) da = \int_4^\infty \left (\frac{x}{2^x} \left ) dx$$ when $$a = x$$
That is, are they not the same?

assuming that both converge.

I'm assuming they do as you have suggested it! But how do I make a safe assumption that this is the case?
Thanks again!

 

[Mark44]

There's the Integral Test for infinite series...

BTW, this is not a Precalculus problem. It should be in the Calculus and Beyond section.

 

[nobahar]

There's the Integral Test for infinite series...

Thanks, I'll look it up. Might need to wish me luck!...

BTW, this is not a Precalculus problem. It should be in the Calculus and Beyond section.

Apologies, but I didn't know that it would entail calculus.

Once again, thanks for all your help!

 

[nobahar]

You could approximate
$$2 \sum_{a=4}^{\infty}\frac{a}{ 2^a}$$
by this integral:
$$2 \int_4^{\infty} \frac{x dx}{2^x}$$

How do I go about doing this? It's difficult to integrate the function. I tried the easy way out by using the calculator but that didn't work.

 

[Mark44]

Change to $$2 \int_4^{\infty} x e^{-x ln 2}dx}$$
This can probably be done by integration by parts, with u = x, and dv = e^{-x ln2}dx. That's how I would start.

 

[Mark44]

Also, since this is an improper integral, you would need to take a limit:
$$2 \lim_{b \rightarrow \infty} \int_4^b x e^{-x ln 2}dx}$$

 

[nobahar]

Thanks Mark, it seems you always come to my rescue.
Do I integrate first with the limit present for b?
i.e. $$\int_4^b x e^{-x ln 2}dx}$$, then employ the limit?
I am not familiar with this, it's more 'advanced' than what I have come across so far.

I'll go away and have a crack at it tomorrow, start afresh. (It's fairly late in Merry Old England.)
Once again, thanks for the continuing help.

 

[Mark44]

Yes, integrate first, which will give you an expression involving b. Then take the limit.

 

[JG89]

lol sorry about that nobahar, my mistake.

 

[nobahar]

lol sorry about that nobahar, my mistake.

No problem!

I attempted to integrate the function:
$$\int xe^{-x \ln{2}} dx = \frac{x * e^{-x \ln{2}}}{- \ln{2}} \left \left - \left \left \frac{e^{-x \ln{2}}}{(\ln{2})^2} = \left - \left \frac{(\ln {2})^2 * x * e^{-x \ln{2}} - e^{-x \ln{2}} * \ln{2}}{(\ln{2})^3} = \left - \left \frac{\ln{2} * x * e^{-x \ln{2}} - e^{-x \ln{2}}}{(\ln{2})^2}$$
(Including +C)
Is this correct so far?

 

[nobahar]

Attempt to 'simplify':


$$\left - \left \frac{e^{-x \ln{2}}(\ln{2} * x + 1)}{(\ln{2})^2}$$

 

[Mark44]

Assuming that you did integration by parts correctly (I didn't check), and that you ended up with this, about the only thing I can see to do to simplify it is replace e^-xln2 with 1/2^x.

 

[nobahar]

I'll post the workings.

$$\int xe^{-x \ln{2}} dx$$

$$u = x \left \left and \left \left \frac{dv}{dx} = e^{-x \ln{2}} \left \left then \left \left \frac{du}{dx} = 1 \left \left and \left \left v = \int e^{-x \ln{2}} dx = \frac{e^{-x \ln{2}}}{- \ln{2}}$$

$$\int xe^{-x \ln{2}} dx = \frac{xe^{-x \ln{2}}}{- \ln{2}} \left \left - \left \left \int \frac{e^{-x \ln{2}}}{- \ln{2}} dx = \frac{xe^{-x \ln{2}}}{- \ln{2}} \left \left - \left \left \frac{e^{-x \ln{2}}}{(\ln{2})^2}$$

 

[nobahar]

Okay, so:
$$\frac{x \ln{2} + 1}{2^x(\ln{2})^2}$$

Now I need:
$$\lim_{b \rightarrow \infty}[\frac{x \ln{2} + 1}{2^x(\ln{2})^2}]^{x=b}_{x=4}]$$

I don't know how to do this. Where would I start?

 

[Tedjn]

It seems to me that JG89's comment is in the right direction of the usual way to work this problem. It does involve a trick that you'll need to see spelled out in detail. Remember it and use it to solve similar problems. Consider for some positive integer $n$ the function

$$f(x) = \frac{x^n}{2^n}$$​

It is everywhere continuous and differentiable, and its derivative is

$$f'(x) = \frac{n}{2^n}x^{n-1}$$​

Next, consider the sum:

$$F_n(x) = \sum_{a=4}^n \frac{x^a}{2^a}$$​

so that we have a geometric series, i.e.

$$F_n(x) = \sum_{a=4}^n \frac{x^a}{2^a} = \frac{x^4/16 - x^5/2^{n+1}}{1-x/2}$$​

The final step is to take the derivative:

$$F'_n(x) = \sum_{a=4}^n \frac{a}{2^a}x^{a-1} = \frac{(x^3/4-5x^4/2^{n+1})(1-x/2) + 1/2(x^4/16-x^5/2^{n+1})}{(1-x/2)^2}$$​

Notice that

$$\lim_{n \rightarrow \infty} F'_n(1) = \sum_{a=4}^\infty \frac{a}{2^a} = 5/4$$​

is the limit that you are looking for.

 

[nobahar]

That's truly amazing. Many, many thanks Tedjn. Initially, it's a little hard to follow because the derivitive is of the variable x, but then you take the sum of a. But I think I understand it. I'll work through it again.
Once again, many thanks.