Question

Is this sum convergent? If so, find the sum.
sum_(n=1)^infinity (2^n+n^2+n)/(2^(n+1)n(n+1))

Answer 1

\[\sum_{n=1}^{\infty} {\frac{2^n+n^2+n}{2^{n+1}n(n+1)}} \]
is that right?

Answer 2

Absolutely

Answer 3

You can divide it into sum of 2 sums :
sum 1 : \[\sum_{n=1}^{\infty} {\frac{1}{2n(n+1)}}\]
sum 2 : \[\sum_{n=1}^{\infty} {\frac{1}{2^{n+1}}}\]

Answer 4

converging since the limit of last term ,when n tends to infinity is infinity..

Answer 5

Thanks anhkhoavo1210, I see how you got that. What would be the next step?

Answer 6

You see the sum 1 can be divided into :
\[\frac{1}{2}(\sum_{n=1}^{\infty} {\frac{1}{n}}-\sum_{n=1}^{\infty} {\frac{1}{n+1}}) \]
So you can conclude these sums converge or not?

Answer 7

Well \[\sum_{n+1}^{\infty} {1 \over n}\] is a harmonic series, which is divergent, correct? I am confused on operating with sums/differences of divergent series, though.

Answer 8

typo: n+1 should be n=1 in the series notation.

Answer 9

you know \[\sum_{n=1}^{\infty} {\frac{1}{n+1}}\] can be replaced into \[\sum_{m=2}^{\infty} {\frac{1}{m}}\], is the form of harmonic series?

Answer 10

I do not really understand that concept. Can you please elaborate?

Answer 11

put \[m=n+1\]

Answer 12

Ah, I see. That makes sense.

Answer 13

But you know :
\[\sum_{n=1}^{\infty} {\frac{1}{n}}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\]
\[\sum_{m=2}^{\infty} {\frac{1}{m}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\]
If you use the subtraction here, you have sum 1 = 1/2
That's so surprise. :)

Answer 14

Ok, I see.

Answer 15

Correct me if I am wrong:
the second series is a geometric series with R = 1/2. Hence the sum of the second series is also 1/2. The sum of the series 1 and series 2 are therefore 1 and the entire series is convergent?

Answer 16

Of course, the entire series is convergent, but I try to calculate sum 2 x)
That is a problem of technic...
\[\sum_{k=0}^{\infty} {x^k}=1+x+x^2+...=\frac{1}{1-x}, |x|<1\]
I try to find similarities of the above form and sum 2.
sum 2 is equivalent to \[\sum_{m=2}^{\infty} {\frac{1}{2^m}}\] , using m=n+1 again
Take x=1/2 into the form. The form =
\[\sum_{k=0}^{\infty} {(1/2)^k}=1+(1/2)+(1/2)^2+...=1+(1/2)+\sum_{m=2}^{\infty} {\frac{1}{2^m}}\]
Then
\[\frac{1}{1-(1/2)}=1+(1/2)+\sum_{m=2}^{\infty} {\frac{1}{2^m}}\]
Yup it equals to 1/2.

Answer 17

Thank you!
Another question: is there a way to tell from the beginning that the series is convergent?

Answer 18

sr I don't understand sth what you say. The series here is general?

Answer 19

How can you tell that the series converges before determining sum?

Answer 20

of course, use limit and other tests like root test and all..

Answer 21

In this series above, because you are not sure the convergence of sum 1, you must try to calculate it.

Answer 22

Awesome. Thanks a lot!