Question

How do you solve `sum_(n=1)^oo (x+1)^n/(n^2+n+2) = 2` ?

Answer 1

No real solutions.

Explanation:

`sum_(k=1)^oo y^k/(k^2+k+2) le sum_(k=1)^oo y^k/(k(k+1))`

`sum_(k=1)^oo y^k/(k(k+1))=1/2(sum_(k=1)^oo y^k/k-sum_(k=1)^oo y^k/(k+1))`

now if `abs y < 1`

`sum_(k=0)^oo y^k = 1/(1-y)` and

`sum_(k=1)^oo y^k/k=int (dy/(1-y))=-log_eabs(1-y)`

also

`sum_(k=1)^oo y^k/(k+1)=1/y sum_(k=2)^oo y^k/k`

and

`sum_(k=2)^oo y^k/k=int dy/(1-y)-y`

so

`sum_(k=1)^oo y^k/(k(k+1))=(-log_e(1-y)-1/y(-log_e (1-y)-y)) = 1/(y)(y+(1-y)log_eabs(1-y))`

or making `y = x+1`

`sum_(k=1)^oo (x+1)^k/(k(k+1)) = (1 + x - x log_eabs[x])/( (1 + x))`

We know that

`sum_(k=1)^oo y^k/(k^2+k+2) < (1 + x - x log_eabs[x])/( (1 + x))`

and

`max_x ( (1 + x - x log_eabs[x])/( (1 + x)) )approx 1 < 39/4`

so for `abs(x+1) < 1` the equation

`sum_(k=1)^oo (x+1)^k/(k^2+k+2)=39/4` has not real solutions.

Answer 2

There is no real value of `x` satisfying the given equation.

Explanation:

First note that:

`sum_(n=1)^N 1/(n(n+1)) = sum_(n=1)^N (1/n-1/(n+1))`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = sum_(n=1)^N 1/n-sum_(n=1)^N 1/(n+1)`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = sum_(n=1)^N 1/n-sum_(n=2)^(N+1) 1/n`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = 1+color(red)(cancel(color(black)(sum_(n=2)^N 1/n)))-color(red)(cancel(color(black)(sum_(n=2)^N 1/n))) - 1/(N+1)`

`color(white)(sum_(n=1)^N 1/(n(n+1))) = 1-1/(N+1)`

So:

`sum_(n=1)^oo 1/(n(n+1)) = lim_(N->oo) (1 - 1/(N+1)) = 1`

Then:

`n^2+n+2 > n^2+n = n(n+1)`

So:

`1/(n^2+n+2) < 1/(n(n+1))" "AA n > 0`

So:

`sum_(n=1)^oo 1/(n^2+n+2) < sum_(n=1)^oo 1/(n(n+1)) = 1`

So if `abs(x+1) <= 1` then:

`abs(sum_(n=1)^oo (x+1)^n/(n^2+n+2)) <= sum_(n=1)^oo 1/(n^2+n+2) < 1`

`color(white)()`
Further note that:

`lim_(n->oo) (1/((n+1)^2+(n+1)+2))/(1/(n^2+n+2)) = 1`

Hence if `abs(x+1) > 1` then `sum_(n=1)^oo (x+1)^n/(n^2+n+2)` diverges by the ratio test.

So for any real value of `x` the sum either diverges or has value with absolute value less than `1`.

So there are no real solutions to the original problem.