Prove that for $vert xvert le 1$ $sum_{n=1}^{infty} frac {sin n}{n} x^n=frac 12 i left(ln (1-xe^i)-...
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Prove that for $vert xvert lt 1$ $$sum_{n=1}^{infty} frac {sin n}{n} x^n=frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
My try:
I solved this using the fact that $$sin x=frac {e^{ix}-e^{-ix}}{2i}$$
Hence $$sum_{n=1}^{infty} frac {sin n}{n} x^n=sum_{n=1}^{infty} frac {e^{in} -e^{-in}}{2in}=frac {-i}{2}left(sum_{n=1}^{infty} frac {(xe^i)^n. }{n}-sum_{n=1}^{infty} frac {(xe^{-i})^n}{n}right) =frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
I wanted to know if there could be any other alternative approach.
calculus sequences-and-series trigonometry
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
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add a comment |
$begingroup$
Prove that for $vert xvert lt 1$ $$sum_{n=1}^{infty} frac {sin n}{n} x^n=frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
My try:
I solved this using the fact that $$sin x=frac {e^{ix}-e^{-ix}}{2i}$$
Hence $$sum_{n=1}^{infty} frac {sin n}{n} x^n=sum_{n=1}^{infty} frac {e^{in} -e^{-in}}{2in}=frac {-i}{2}left(sum_{n=1}^{infty} frac {(xe^i)^n. }{n}-sum_{n=1}^{infty} frac {(xe^{-i})^n}{n}right) =frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
I wanted to know if there could be any other alternative approach.
calculus sequences-and-series trigonometry
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
$endgroup$
add a comment |
$begingroup$
Prove that for $vert xvert lt 1$ $$sum_{n=1}^{infty} frac {sin n}{n} x^n=frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
My try:
I solved this using the fact that $$sin x=frac {e^{ix}-e^{-ix}}{2i}$$
Hence $$sum_{n=1}^{infty} frac {sin n}{n} x^n=sum_{n=1}^{infty} frac {e^{in} -e^{-in}}{2in}=frac {-i}{2}left(sum_{n=1}^{infty} frac {(xe^i)^n. }{n}-sum_{n=1}^{infty} frac {(xe^{-i})^n}{n}right) =frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
I wanted to know if there could be any other alternative approach.
calculus sequences-and-series trigonometry
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
$endgroup$
Prove that for $vert xvert lt 1$ $$sum_{n=1}^{infty} frac {sin n}{n} x^n=frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
My try:
I solved this using the fact that $$sin x=frac {e^{ix}-e^{-ix}}{2i}$$
Hence $$sum_{n=1}^{infty} frac {sin n}{n} x^n=sum_{n=1}^{infty} frac {e^{in} -e^{-in}}{2in}=frac {-i}{2}left(sum_{n=1}^{infty} frac {(xe^i)^n. }{n}-sum_{n=1}^{infty} frac {(xe^{-i})^n}{n}right) =frac 12 i left(ln (1-xe^i)- ln(1-xe^{-i})right) $$
I wanted to know if there could be any other alternative approach.
calculus sequences-and-series trigonometry
calculus sequences-and-series trigonometry
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
edited Dec 8 '18 at 11:37
Digamma
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
asked Dec 8 '18 at 11:32
DigammaDigamma
6,1621440
6,1621440
add a comment |
add a comment |
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