Dear All, could you help me to solve following?

I need to know, how much i is and how to calculate it.

Thanks milion times

\(\displaystyle (1+i)^{-60}\) 60 times is just \(\displaystyle 60(1+i)^{-60}=39.55\ or\ =3,955\)

so, assuming the sum should be from k=1 to 60

and for convenience letting x=1+i, solving for x,

when finally i=x-1

\(\displaystyle \frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+....+\frac{1}{x^{60}}\)

\(\displaystyle =\frac{1}{x}\left(1+\frac{1}{x}+\frac{1}{x^2}+.....+\frac{1}{x^{59}}\right)\)

is a geometric series and equals \(\displaystyle \frac{1}{x}\left(\frac{\frac{1}{x^{60}}-1}{\frac{1}{x}-1}\right)=\frac{\frac{1}{x^{60}}-1}{x(1-x)}\)

Solving this graphically, using root-finding software, we get

\(\displaystyle \frac{\frac{1}{x^{60}}-1}{x(1-x)}-3,955=0\)

for x=0.907783 approximately.

Slight variations in x yield quite different results.

This gives i=x-1=0.907783-1=-0.092217

If the sum of the series is 39.55 rather than 3,955,

this yields x=1.0143, so i=0.0143 approximately.