Question

How do you find the power series for `f(x)=int ln(1+t)/tdt` from [0,x] and determine its radius of convergence?

Answer 1

`f(x) = x - x^2/4 + x^3/9 - x^4/16 +...`
`" " = sum_(r=1)^oo (-1)^(r+1) x^r/r^2`

radius of convergence is `|x| lt 1`

or equivalently `x in (-1,1) " or " -1 lt x lt 1`

Explanation:

We start with the well known Maclaurin Series for `ln(1+x)`:

`ln (1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...`

If we replace `x` in the series by `t` then we get:

`ln (1+t) = t - t^2/2 + t^3/3 - t^4/4 +...`

So then we have:

`ln (1+t)/t = 1/t{ t - t^2/2 + t^3/3 - t^4/4 +... }`
`" " = 1 - t/2 + t^2/3 - t^3/4 +...`

So:

`f(x) = int_0^x \ ln(1+t)/t \ dt`
`" " = int_0^x \ {1 - t/2 + t^2/3 - t^3/4 +... } \ dt`
`" " = [ t - (t^2/2)/2 + (t^3/3)/3 - (t^4/4)/4 +... ]_0^x`
`" " = [ t - t^2/4 + t^3/9 - t^4/16 +... ]_0^x`
`" " = { x - x^2/4 + x^3/9 - x^4/16 +... } - { 0 }`
`" " = x - x^2/4 + x^3/9 - x^4/16 +...`

`" " = sum_(r=1)^oo (-1)^(r+1) x^r/r^2`

Radius of Convergence

It is probably reasonable to expect the ROC of this result is the same as that of `ln(1+x)`, which is `|x| lt 1`, but we can derive the actual result to be sure

We can apply d'Alembert's ratio test:

Suppose that;

`S=sum_(r=1)^oo a_n \ \`, and `\ \ L=lim_(n rarr oo) |a_(n+1)/a_n|`

Then

if L < 1 then the series converges absolutely;
if L > 1 then the series is divergent;
if L = 1 or the limit fails to exist the test is inconclusive.

So our series is;

`S = sum_(r=1)^oo (-1)^(r+1) x^r/r^2`

So our test limit is:

`L = lim_(n rarr oo) | { ( (-1)^(n+1+1) x^(n+1) ) / ( (n+1)^2 ) } / { ( (-1)^(n+1) x^n ) / n^2 } |`
`\ \ \ = lim_(n rarr oo) | ( (-1) x n^2 ) / (n+1)^2 |`
`\ \ \ = lim_(n rarr oo) |-x | xx | ( n^2 ) / (n^2+2n+1) * (1/n^2)/(1/n^2) |`
`\ \ \ = |x| \ lim_(n rarr oo) | ( 1 ) / (1+2/n+1/n^2) |`
`\ \ \ = |x|`

And we can conclude that the series converges if `L lt 1`

`=> |x| lt 1`. as predicted