Infinite series convergence/divergence and finding the sum?

4/n²+4n+3

4/(n+1)(n+3)

4Σ(with n from 1 to ∞) 1/(n+1)(n+3)

this is a telescoping series:

1/(n+1)(n+3) =A/(n+1) +B/(n+3)

1=A(n+3)+B(n+1)

n=-3:

1=B(-2)

-2B=1 =>B=-1/2

n=-1:

1=2A =>A=1/2

4Σ 1/2(n+1) -1/2(n+3)

2Σ 1/n+1 -1/n+3

n=1,2,3,4,etc

[1/2-1/4]+[1/3-1/5]+[1/4-1/6]+[1/5-1/8]+[-1/k+2]+[-1/k+4]

except from 1/2 and 1/3 the rest of the terms will cancel.

1/2 +1/3 -1/k+2 -1/k+4

as n→ ∞

lim [ 1/2+1/3 -1/k+2-1/k+4] = 5/6

originally we had 2Σ 1/n+1 -1/n+3

the series converges to 2(5/6) = 10/6

For n ≥ 1, n^2 + 4n + 3 > n^2 > 0, so we have:

0 < 1/(n^2 + 4n + 3) < 1/n^2 ==> 0 < 4/(n^2 + 4n + 3) < 4/n^2.

Since ∑ 4/n^2 (from n=1 to infinity) converges because it is a multiple of a convergent p-series, the series in question converges by the Comparison Test.

Now, to find the sum of this series, notice that n^2 + 4n + 3 can be nicely factored to (n + 3)(n + 1), so we can find the sum by using partial fractions. For some A and B, we have:

4/[(n + 3)(n + 1)] = A/(n + 3) + B/(n + 1).

Multiplying both sides by (n + 3)(n + 1) to clear the fractions gives:

4 = A(n + 1) + B(n + 3) = (A + B)n + (A + 3B).

Comparing the coefficients of n (note that the coefficient of n on the left side is 0) gives:

A + B = 0 ==> A = -B.

Then, comparing the constants gives:

A + 3B = 4,

but since A = -B, this becomes:

-B + 3B = 4 ==> B = 2.

Using this value of B and the fact that A = -B, A = -2. So, the required partial fraction decomposition is:

4/[(n + 3)(n + 1)] = 2/(n + 1) - 2/(n + 3).

Now, let us consider the sum of the first k terms of the series. Using the above decomposition:

∑ 4/(n^2 + 4n + 3) (from n=1 to k) = 2 ∑ [1/(n + 1) - 1/(n + 3)] (from n=1 to k)

= 2{[1/2 + 1/3 + 1/4 + ... + 1/(k + 1)] - [1/4 + 1/5 + 1/6 + ... + 1/(k + 3)]}

= 2[1/2 + 1/3 - 1/(k + 1) - 1/(k + 2)], as all other terms cancel in pairs.

Letting k --> infinity yields the required sum to be 2(1/2 + 1/3 - 0 - 0) = 5/3.

I hope this helps!

Compare with 4/n^2

Since 4/n^2 converges ( p-series) then 4/((n^2)+4n+3)

converges.

To find the sum use

4/((n^2)+4n+3) = 2 /(n+1) - 2/(n+3)

the series converges let's prove it

n^2 +4n +3

=n^2+4n +4 -1

=(n+2)^2-1

=(n+1)(n+3)

so the nth term is

4/[(n+1)(n+3)]

taking a partial fraction expansion you get

2/(n+1) - 2/(n+3)

this is now a telescoping series and the only terms which are not canceled are 2/2+2/3

(convince yourself: 2/2-2/4+2/3-2/5+2/4-2/6+2/5-2/7.........)

and therefore the sum is equal to 2/2+2/3 = 5/6

hope this helps

considering the fact that I spent the time working in this subject... This series converges by way of the Alternating series try. (i) {(sqrt(n))sin(a million/n)} is reducing for sufficiently super n. that is maximum only considered by way of Calculus. permit f(x) = x^(a million/2) sin(a million/x). ==> f'(x) = (x^(-3/2)/2) [x sin(a million/x) - 2 cos(a million/x)]. As x-->infinity, x sin(a million/x) - 2 cos(a million/x) = sin(a million/x) / (a million/x) - 2 cos(a million/x) --> a million - 2 = -a million. So, f'(x) < 0 for sufficiently super n. for this reason f is reducing for sufficiently super n. (ii) lim(n-->infinity) n^(a million/2) sin(a million/n) = lim(n-->infinity) n^(-a million/2) * [sin(a million/n) / (a million/n)] = 0 * a million = 0. ------------------------------------ i desire this helps! fact: i'm no longer too specific if this sum has a closed form. regardless of if, considering the fact that this series is alternating, possible approximate it with the 1st n words with an blunders in magnitude no bigger than the (n+a million)-th term.