Can you explain why this series conditionally convergent?

T(k)=(k+7)/[k(k+4)]=>

T(k+1)=(k+8)/[(k+1)(k+5)]

T(k)-T(k+1)=

(k+7)/[k(k+4)]-(k+8)/[(k+1)(k+5)]=

(k^2+15k+35)/[k(k+1)(k+4)(k+5)]>0

(since k>=1)

=>

T(k)>T(k+1)=>

the terms in the series are decreasing

in the absolute value & T(k)->0 as k->inf.

The series is an alternating decreasing

series=>it converges conditionally.

(You can find the theorem in a math-book

supporting this conclusion, the Leibnitz's Test ).

|T(k+1)/T(k)|=k(k+8)(k+4)/[(k+1)(k+5)(k+7)]=>

|T(k+1)/T(k)|-1=

-(k^2+15k+35)/(k^3+13k^2+47k+35)

ln(k)[-(k^2+15k+35)/(k^3+13k^2+47k+35)]=

-ln(k)/[(k^3+13k^2+47k+35)/(k^2+15k+35)]=g(k)

limit g(k)=

k->inf.

limit { -1/{k[(k^4+21k^3+...-35)]/(k^2+15k+35)^2}}

k->inf.

=0=>the series diverges absolutely.

This test is known as Bertrand's Test.

(i) Σ(k=1 to ∞) (-1)^(k+1) (k+7)/(k(k+4)) converges by the Alternating Series Test, since {(k+7)/(k(k+4))} is a decreasing sequence which converges to 0.

To show the 'decreasing' part, note that

(d/dk) (k+7)/(k(k+4))

= (d/dk) (k+7)/(k^2+4k)

= [(k^2+4k) - (k+7)(2k+4)] / (k^2+4k)^2

= -(k^2+14k+28) / (k^2+4k)^2

< 0 for all k > 0.

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(ii) Σ(k=1 to ∞) |(-1)^(k+1) (k+7)/(k(k+4))|

= Σ(k=1 to ∞) (k+7)/(k(k+4)) diverges by the Comparison Test.

To show this, note that for all k > 0:

(k+7)/(k(k+4)) > (k+4)/(k(k+4)) = 1/k, and Σ(k=1 to ∞) 1/k is the divergent p-series.

Hence, Σ(k=1 to ∞) (-1)^(k+1) (k+7)/(k(k+4)) is conditionally convergent.

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I hope this helps!