Question

Determine if the series is convergent or divergent, and please explain how you came to get your answers. Thanks!

a. 20-15+10-5...

b. 12/5 + 48/25 + 192/125...

c. -8/5 + 32/25 - 128/125 + ...

d. 2 + 4/2^2 + 8/3^2 + 16/4^2...

Answer 1

b) \[\sum_{n}^{\infty}3(4/5)^{n} \] = \[3((1-4/5^\infty)/(1-4/5))= 3(1/1-4/5)=15\]

Answer 2

so b) converges

Answer 3

Thank you so much! The teacher showed these as samples today but didn't explain how she got her answers!

Answer 4

I've always struggled with the ones that oscillate up and down like a) and c). I'll take a crack at them.
I'm sure it involves a (-1)^n, and that's what makes it bounce back and forth from positive to negative. Check back in awhile.

Answer 5

For a) , I think this is \[\sum_{n=1}^{\infty} 20 + (-1)^n(a _{n}-5)\]
alternatively, you could group write out several more numbers and see what happens so 20-15+10-5+0-(-5)+(-10)-(-15)+(-20)....

Either way this series diverges (goes to larger and larger number)

Answer 6

@Julian101 The formula
\[ \sum_{n=1}^{k} ar^n = \frac{a(1-r^k)}{1-r}\] can only be used if \(k\) is finite. If you have \(k=\infty\), you should use
\[ \sum_{n=1}^{\infty} ar^n = \frac{ar}{1-r},\] which if you apply to the series in question:
\[ \sum_{n=1}^{\infty} 3\left( \frac{4}{5} \right)^n = \frac{3(4/5)}{1-(4/5)} = 12.\] In short, your answer included an extra term \(3(4/5)^0 = 3\). And since I'm a pedant, it should also be pointed out that writing infinity as a power is never good practice! :)

Answer 7

Erin,
Thanks for your correction. I've always struggled with these. If you'd like, have a look at a), to see if I've gone wrong anywhere.

Answer 8

Yeah... series are tricky, aren't they? I use to struggle so much over these.
In your answer to (a), I'm not sure what you mean by \(a_n\)?

Answer 9

\[a _{n}\] was my nth term of the sequence. So \[a _{1}\] is 20, \[a _{2}\] is 15, etc.

Answer 10

Oh I see. I think I would write the series as

20-15+10-5+0-(-5)+(-10)-(-15)+(-20)-(-25)+(-30)-...,

and notice that the first 9 terms cancel, so the series is just 25-30+35-40+..., which makes it a bit easier to deal with! So the series is
\[\sum_{n=1}^{\infty} 5(-1)^{n+1} (n+4).\]In any case, you're definitely right that it diverges.

Answer 11

Ah, I see!
Now...is this a geometric series, or an arithmetic?
I'm leaning towards geometric now that I see your formula.

Answer 12

Would your first term start at n=0, though? Otherwise wouldn't the first term be 25?

Answer 13

I want the first term to be 25, because the series (after I simplified it) is:
25-30+35-40+...
It's neither really. For it to be a geometric, I need to multiply every term by the same number to get the next term, and you can see that whatever I multiply 25 by to get -30 is not the same as what I multiply -30 by to get 35.
It's also not arithmetic, since the difference between 25 and -30 is not the same as the difference between -30 and 35.

Answer 14

OK. These series that involve alternating signs almost seem like they are in a class all of their own.

Answer 15

Indeed they are, and they're called alternating series! They make things more interesting, no? :D

Answer 16

They drive me nuts, because they seem like they should be easy to figure out.Then I look up and an hour has evaporated, while I'm still stuck!

Would you like to do the remaining two questions for the original poster of these problems? It would most likely be more beneficial for her.

Answer 17

Well, you were intending to do (c), weren't you? Why don't you try that one and I'll take a crack at (d)?

Answer 18

It's a deal!

Answer 19

c) is -8/5 + 32/25-128/125...

\[\sum_{n=1}^{\infty} 2(-1)^n(4/5)^n\]

which should converge, since 4/5 should eventually go to zero as n goes to infinity

Answer 20

Agreed. This one is a geometric series, since you can write it as
\[\sum_{n=1}^{\infty} 2\left( -\frac{4}{5} \right)^n\]and so using the formula from before, you can find what it converges to.

As for (d), I'm assuming the power applies to the bottom only, so the series is
\[\frac{2}{1^2} + \frac{4}{2^2} + \frac{8}{3^2} + \frac{16}{4^2} + \dots = \sum_{n=1}^{\infty} \frac{2^n}{n^2},\] which you can show diverges by the limit test.

Answer 21

Thanks again for your help. A very illuminating "series" of problems!