Approximating and Sorting Values of Series

[Mampac]

Howdy?
I'm facing the following problem:
For alpha, I write out all the terms and see that they result in cancelations, and only the first and last terms remain. Computing their limit, the last term's value is 0, and the first one's (2/(1 + 1)) is 1 so I can say alpha equals to 1.

For beta, I use the formula for sum of geometric progression where |q| < 1: I do (5/7)/(1 - 5/7) and get 2/5 so beta equals to 2/5.

For gamma (or which letter is it?), I have no idea what to do. This is an alternating series and I see no way of evaluating this. One way of evaluating such stuff is when they look like some famous Taylor series, but this one doesn't seem to correspond to any of these. A classmate told me to "check the absolute convergence" but I fail to relate it to this.

For delta, I once again can't think of any way to evaluate this. This is no geometric series as beta.

Thank you in advance,
Have a nice day.

 

[lex]

[MATH]\gamma[/MATH] is clearly negative and [MATH]\delta[/MATH] is undefined (since n=0 is included)

For beta, I use the formula for sum of geometric progression where |q| < 1: I do (5/7)/(1 - 5/7) and get 2/5 so beta equals to 2/5.

There is an arithmetic error here and the sequence starts with n=0, so you are doing the incorrect calculation too. The sum is [MATH]\frac{7}{2}[/MATH]

 

[Mampac]

[MATH]\gamma[/MATH] is clearly negative and [MATH]\delta[/MATH] is undefined (since n=0 is included)


There is an arithmetic error here and the sequence starts with n=0, so you are doing the incorrect calculation too. The sum is [MATH]\frac{7}{2}[/MATH]

can you justify gamma being negative, please?
i also noticed that 1/0 is undefined for delta. do you think it's a typo? if not, how can i then compare it with other numbers? there's 2 options where delta stands as the smallest number, does that mean it's one of them?

thanks for catching the mistake, my bad!

 

[lex]

The ratio test shows that the series for [MATH]\gamma[/MATH] is absolutely convergent.
So, it is ok to group the alternating sequence and it will still converge to [MATH]\gamma[/MATH]:
[MATH]\left(-1 + \frac{1}{6!}\right) + \left(-\frac{1}{11!} + \frac{1}{16!}\right) + \left(-\frac{1}{21!} + \frac{1}{26!}\right)+...[/MATH]Each bracket is clearly negative, therefore the sum is negative.

As for the final series starting with n=1.
I note that [MATH]\frac{1}{1-x}[/MATH] is [MATH]\hspace2ex 1 + x + x^2 + x^3[/MATH] ...= [MATH]\sum_{n=0}^{n=\infty}x^n[/MATH]Integrating we get [MATH]\hspace1ex \ln \left(\frac{1}{1-x} \right)= x + \frac{x^2}{2} + \frac{x^3}{3} + ... = \left(\sum_{n=0}^{n=\infty}\frac{x^{n+1}}{n+1}\right)[/MATH] (taking x<1)
Evaluate this at [MATH]x=\frac{1}{2}[/MATH] and we get [MATH]\hspace1ex \ln(2) = \left(\tfrac{1}{2}\right) + \frac{(\tfrac{1}{2})^2}{2} + \frac{\left(\tfrac{1}{2}\right)^3}{3} + ... = \sum_{n=1}^{n=\infty}\frac{1}{n2^n}[/MATH]
So [MATH]\alpha[/MATH] = 1, [MATH]\beta[/MATH] = [MATH]\frac{7}{2}[/MATH], [MATH]\gamma[/MATH] < 0, [MATH]\delta[/MATH] undefined (starting at n=1, it is ln(2))

 

[Mampac]

The ratio test shows that the series for [MATH]\gamma[/MATH] is absolutely convergent.
So, it is ok to group the alternating sequence and it will still converge to the [MATH]\gamma[/MATH]:
[MATH]\left(-1 + \frac{1}{6!}\right) + \left(-\frac{1}{11!} + \frac{1}{16!}\right) + \left(-\frac{1}{21!} + \frac{1}{26!}\right)+...[/MATH]Each bracket is clearly negative, therefore the sum is negative.

As for the final series starting with n=1.
I note that [MATH]\frac{1}{1-x}[/MATH] is [MATH]\hspace2ex 1 + x + x^2 + x^3[/MATH] ...= [MATH]\sum_{n=0}^{n=\infty}x^n[/MATH]Integrating we get [MATH]\hspace1ex \ln \left(\frac{1}{|1-x|} \right)= x + \frac{x^2}{2} + \frac{x^3}{3} + ... = \left(\sum_{n=0}^{n=\infty}\frac{x^{n+1}}{n+1}\right)[/MATH]Evaluate this at [MATH]x=\frac{1}{2}[/MATH] and we get [MATH]\hspace1ex \ln(2) = \left(\tfrac{1}{2}\right) + \frac{(\tfrac{1}{2})^2}{2} + \frac{\left(\tfrac{1}{2}\right)^3}{3} + ... = \sum_{n=1}^{n=\infty}\frac{1}{n2^n}[/MATH]
So [MATH]\alpha[/MATH] = 1, [MATH]\beta[/MATH] = [MATH]\frac{7}{2}[/MATH], [MATH]\gamma[/MATH] < 0, [MATH]\delta[/MATH] undefined (starting at n=1, it is ln(2))

thanks a lot. it's definitely a typo since there exists a correct answer in that case (option A).

omg, never in my life I'd notice all these similarities in the final series. how am i supposed to solve such things during an exam -- no one knows... i do admit, though, that computation of only first 3 series is enough to select the right option. once again, thank you

 

[Cubist]

I might generate a backlash for saying this, but if you're short of time in an exam you could make an educated guess that gamma is negative. How? You probably know that factorials become big VERY quickly. Therefore the gamma terms become small VERY rapidly after the first "-1". It therefore seems very unlikely that the sum's value would be able to get back towards 0, especially because terms oscillate in sign. However, @lex 's proof is brilliant (and certain) if you can spot it!

 

[lex]

I suppose it's just practice that is the key. But as you say, it's harder doing it in an exam. Plenty of past papers! Good luck.

 

[lex]

I got side-tracked from the fact that this is for an exam. So, you do them without evaluation if possible.
E.g. the final one, just note that: [MATH]n<2^n \Rightarrow \frac{1}{2^n}.\frac{1}{2^n}<\frac{1}{n2^n}<\frac{1}{n}.\frac{1}{n}[/MATH]so [MATH]\frac{1}{4^n}<\frac{1}{n2^n}<\frac{1}{n^2}[/MATH]so [MATH]\frac{1}{3}≤\delta≤\frac{\pi^2}{6}[/MATH] which is sufficient to place [MATH]\delta[/MATH].
(I think it's worth knowing the sum of [MATH]\tfrac{1}{n^2}[/MATH]).

 

[lex]

I should have noted that [MATH]\frac{1}{1-x}\hspace1ex [/MATH] is [MATH]\hspace1ex 1 + x + x^2 + x^3[/MATH] ...= [MATH]\sum_{n=0}^{n=\infty}x^n[/MATH]for |x|<1