Convergent and divergent series

Hey,
I've been working my way through a load of examples on using the comparison test and the ratio test to find out whether a series is convergent or divergent, these are the ones that I couldn't do:

1) sum from k=1 to infinity of k³ /4^k

2) sum from k=1 to infinity of (1+tanh(k³ ))/k²
3) sum from k=1 to infinity of k!/100^4k

4) sum from k=1 to infinity of k!/k^k

If anyone could help me on any of these (or even all if you're feeling brave) i would really apprieciate it, thanks guys!!

Reply 1

Christo

14

3) let x_k=k!/100^4k

lim(k->0) x_k+1/x_k = k+1/100 =1/100

1/100<1 so x_k converges.

Reply 2

Jonny W

8

Let a(k) be the kth term.

(1)
a(k + 1)/a(k)
= (1/4)((k + 1)/k)^3
-> 1/4

Convergent by the ratio test.

(2)
-1 <= tanh(x) <= 1 for all real x.

0 <= (1 + tanh(k^3))/k^2 <= 2/k^2 for any integer k >= 1.

Since

\sum_{k=1}^\infty

2/k^2 converges, so does

\sum_{k=1}^\infty

(1 + tanh(k^3))/k^2 by the comparison test.

(3)
a(k + 1)/a(k)
= (k + 1)/100^4
-> infinity

Divergent by the ratio test.

(4)
a(k + 1)/a(k)
= (k + 1) k^k/(k + 1)^(k + 1)
= k^k/(k + 1)^k
= (k/(k + 1))^k
= ((k + 1)/k)^(-k)
= (1 + 1/k)^(-k)
-> 1/e

Convergent by the ratio test.

Reply 3

wibble...

OP

Thanks for the replies!
Just a quick question about Jonny W's reply, in (i) where did the 1/4 come from? and also in (iii) at the end you put that k+1/100⁴ tends to infinity, is there a rule in this case which proves it or is it ok to simply state it?
Thanks again! :biggrin:

Reply 4

Gaz031

15

wibble...

Thanks for the replies!
Just a quick question about Jonny W's reply, in (i) where did the 1/4 come from? and also in (iii) at the end you put that k+1/100⁴ tends to infinity, is there a rule in this case which proves it or is it ok to simply state it?
Thanks again! :biggrin:

In (i), the 1/4 came from the (4^k)/(4^(k+1)) that you get.
In (iii), (k+1)/(100^4)=k/(100^4)+(1/100^4) must tend to infinity since if a sequence tends to infinity then any positive constant multiple of that sequence will also tend to infinity and adding a constant to a sequence that goes to infinity doesn't affect whether it goes to infinity or not. Both of these can be proved from the definitions of a sequence tending to infinity, have a go!

Reply 5

topcrabcakes

11

The 1/4 actually comes from:

(1/4)((k + 1)/k)^3

because (k + 1)/k --> 1 as k-->∞ so ((k + 1)/k)^3 ---> 1

so (1/4)((k + 1)/k)^3 ---> 1/4 < 1 therefore series is convergent.

Reply 6

wibble...

OP

Oh I see sorry didn't see that, thanks for all the help!!

I've been attempting a few more and have a couple of questions about them:
(same question)

sum of k=1 to infinity 12^k /k!

with this one I know that xⁿ /n! tend to 0 for any x so could i just simply say that 12^k /k! (less than or equal to) 13^k /k! so converges?

sum of k=1 to infinity 1/(log(k+1)) (less than or equal to) 1/logk, and as logk tends to infinity 1/logk tends to 0 therefore the sequence converges.

sum of k=1 to infinity (2k² +1)/(5k³ -4)
not sure about this 1

what do you all think about these? thanks again :redface:

Reply 7

topcrabcakes

11

Remember, we are doing series and not sequences. A sequence may -->0 as n-->∞, but it's series may not eg infinite sum of 1/n

Reply 8

Gaz031

15

sum of k=1 to infinity 12^k /k!

with this one I know that xⁿ /n! tend to 0 for any x so could i just simply say that 12^k /k! (less than or equal to) 13^k /k! so converges?

No, the sequence whose terms you are summing being null is a necessary but not sufficient condition for series convergence. The sequence whose terms you are summing looks sufficiently neat to use the ratio test.

sum of k=1 to infinity 1/(log(k+1)) (less than or equal to) 1/logk, and as logk tends to infinity 1/logk tends to 0 therefore the sequence converges.

As above, the sequence whose terms you are summing being null is a necessary but not sufficient condition for series convergence. Try using the comparison test with the inequality log(k+1)<(k+1).

sum of k=1 to infinity (2k² +1)/(5k³ -4)
not sure about this 1

You can see that it's going to behave like the sum of (2/5)(1/k) which is a positive multiple of the harmonic series and will hence diverge, so try using the comparison test.

Reply 9

topcrabcakes

11

wibble...

Oh I see sorry didn't see that, thanks for all the help!!

I've been attempting a few more and have a couple of questions about them:
(same question)

sum of k=1 to infinity 12^k /k!

with this one I know that xⁿ /n! tend to 0 for any x so could i just simply say that 12^k /k! (less than or equal to) 13^k /k! so converges?

sum of k=1 to infinity 1/(log(k+1)) (less than or equal to) 1/logk, and as logk tends to infinity 1/logk tends to 0 therefore the sequence converges.

sum of k=1 to infinity (2k² +1)/(5k³ -4)
not sure about this 1

what do you all think about these? thanks again :redface:

sum of k=1 to infinity 12^k /k!

we know that:

sum of k=1 to infinity x^k /k!= e^x

so sum of k=1 to infinity 12^k /k! = e¹² and is therefore convergent.

For sum of k=1 to infinity 1/(log(k+1)) use ratio test. |(log k)/(log (k+1))|<1 therefore the infinite sum is convergent.

Reply 10

topcrabcakes

11

(2k²+1)/(5k³-4)>2k²/5k³=2/(5k)

but ∑2/(5k) = (2/5)∑(1/k)

but we know 1/k is divergent so (2/5)∑(1/k) is divergent, so

∑(2k²+1)/(5k³-4) is divergent.

Convergent and divergent series

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