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

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