Summation of series $sum_x=1^inftyfrac-theta^xln(1-theta)$
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How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$
I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.
sequences-and-series
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up vote
-1
down vote
favorite
How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$
I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.
sequences-and-series
2
It is a sum of a geometric series.
â Dr. Sonnhard Graubner
Sep 8 at 12:03
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$
I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.
sequences-and-series
How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$
I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.
sequences-and-series
sequences-and-series
asked Sep 8 at 12:02
Mathxx
3,37311131
3,37311131
2
It is a sum of a geometric series.
â Dr. Sonnhard Graubner
Sep 8 at 12:03
add a comment |Â
2
It is a sum of a geometric series.
â Dr. Sonnhard Graubner
Sep 8 at 12:03
2
2
It is a sum of a geometric series.
â Dr. Sonnhard Graubner
Sep 8 at 12:03
It is a sum of a geometric series.
â Dr. Sonnhard Graubner
Sep 8 at 12:03
add a comment |Â
2 Answers
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Well
$$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$
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Just because of the formula
$$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Well
$$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$
add a comment |Â
up vote
3
down vote
accepted
Well
$$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Well
$$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$
Well
$$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$
answered Sep 8 at 12:08
b00n heT
8,90211833
8,90211833
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up vote
2
down vote
Just because of the formula
$$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$
add a comment |Â
up vote
2
down vote
Just because of the formula
$$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Just because of the formula
$$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$
Just because of the formula
$$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$
answered Sep 8 at 12:06
Bernard
112k635104
112k635104
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add a comment |Â
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2
It is a sum of a geometric series.
â Dr. Sonnhard Graubner
Sep 8 at 12:03