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Convert fraction to an infinite series with alternating sign
Testing if a geometric series converges by taking limit to infinityHow does the sum of an infinite series calculated?Eulers Doubly Infinite Geometric SeriesSufficient condition for an infinite series to be zeroSums of infinite seriesAlgebraic numbers and geometric series - from finite to infinite, similarity with transcendental numbersEvaluating an infinite series using partial fractions…value of the infinite serieshelp with sum of infinite series, stuck in problemDifferential Equation Involving Multiplication of Infinite Series?
$begingroup$
$$
f(m)=sum_n=0^inftyBigl(frac-1mBigr)^n\
$$
converges at $fracmm+1$ where $1 < m < infty$. In other words, where $m$ approaches $1$ it converges at $0.5$ and where m approaches $infty$ it converges at $1$.
Does that mean that I can multiply any real number $m$ with $2(f(m)-0.5)$ to converge at that number? It makes sense algebraically, although I might as well just multiply $m$ by $1$, but with an infinite series, $m$ should surely be factored into every term?
Is there maybe another way to convert a fraction into a variation of a geometric series?
sequences-and-series
asked Mar 12 at 9:43
dataphiledataphile
3791310
$endgroup$
add a comment |
$begingroup$
$$
f(m)=sum_n=0^inftyBigl(frac-1mBigr)^n\
$$
converges at $fracmm+1$ where $1 < m < infty$. In other words, where $m$ approaches $1$ it converges at $0.5$ and where m approaches $infty$ it converges at $1$.
Does that mean that I can multiply any real number $m$ with $2(f(m)-0.5)$ to converge at that number? It makes sense algebraically, although I might as well just multiply $m$ by $1$, but with an infinite series, $m$ should surely be factored into every term?
Is there maybe another way to convert a fraction into a variation of a geometric series?
sequences-and-series
asked Mar 12 at 9:43
dataphiledataphile
3791310
$endgroup$
1
$begingroup$
Huh? Your series doesn't converge at all. The terms don't approach zero.
$endgroup$
– Gerry Myerson
Mar 12 at 10:11
$begingroup$
See: wolframalpha.com/input/?i=%5Csum(-1%2Fm)%5En and wolframalpha.com/input/?i=%5Csum(-1%2F3)%5En
$endgroup$
– dataphile
Mar 12 at 10:13
$begingroup$
sorry, I've added brackets now
$endgroup$
– dataphile
Mar 12 at 10:14
2
$begingroup$
OK, now you have a geometric series with first term $-1/m$ and common ratio $-1/m$, so its sum is $-1/(m+1)$, not $m/(m+1)$. Perhaps you want to start with $n=0$ instead of $n=1$?
$endgroup$
– Gerry Myerson
Mar 12 at 10:17
$begingroup$
Cool, thanks. I've fixed n.
$endgroup$
– dataphile
Mar 12 at 10:20
add a comment |
$begingroup$
$$
f(m)=sum_n=0^inftyBigl(frac-1mBigr)^n\
$$
converges at $fracmm+1$ where $1 < m < infty$. In other words, where $m$ approaches $1$ it converges at $0.5$ and where m approaches $infty$ it converges at $1$.
Does that mean that I can multiply any real number $m$ with $2(f(m)-0.5)$ to converge at that number? It makes sense algebraically, although I might as well just multiply $m$ by $1$, but with an infinite series, $m$ should surely be factored into every term?
Is there maybe another way to convert a fraction into a variation of a geometric series?
sequences-and-series
asked Mar 12 at 9:43
dataphiledataphile
3791310
$endgroup$
$$
f(m)=sum_n=0^inftyBigl(frac-1mBigr)^n\
$$
converges at $fracmm+1$ where $1 < m < infty$. In other words, where $m$ approaches $1$ it converges at $0.5$ and where m approaches $infty$ it converges at $1$.
Does that mean that I can multiply any real number $m$ with $2(f(m)-0.5)$ to converge at that number? It makes sense algebraically, although I might as well just multiply $m$ by $1$, but with an infinite series, $m$ should surely be factored into every term?
Is there maybe another way to convert a fraction into a variation of a geometric series?
sequences-and-series
sequences-and-series
asked Mar 12 at 9:43
dataphiledataphile
3791310
asked Mar 12 at 9:43
dataphiledataphile
3791310
edited Mar 12 at 11:27
dataphile
asked Mar 12 at 9:43
dataphiledataphile
3791310
asked Mar 12 at 9:43
dataphiledataphile
3791310
asked Mar 12 at 9:43
dataphiledataphile
3791310
3791310
1
$begingroup$
Huh? Your series doesn't converge at all. The terms don't approach zero.
$endgroup$
– Gerry Myerson
Mar 12 at 10:11
$begingroup$
See: wolframalpha.com/input/?i=%5Csum(-1%2Fm)%5En and wolframalpha.com/input/?i=%5Csum(-1%2F3)%5En
$endgroup$
– dataphile
Mar 12 at 10:13
$begingroup$
sorry, I've added brackets now
$endgroup$
– dataphile
Mar 12 at 10:14
2
$begingroup$
OK, now you have a geometric series with first term $-1/m$ and common ratio $-1/m$, so its sum is $-1/(m+1)$, not $m/(m+1)$. Perhaps you want to start with $n=0$ instead of $n=1$?
$endgroup$
– Gerry Myerson
Mar 12 at 10:17
$begingroup$
Cool, thanks. I've fixed n.
$endgroup$
– dataphile
Mar 12 at 10:20
add a comment |
1
$begingroup$
Huh? Your series doesn't converge at all. The terms don't approach zero.
$endgroup$
– Gerry Myerson
Mar 12 at 10:11
$begingroup$
See: wolframalpha.com/input/?i=%5Csum(-1%2Fm)%5En and wolframalpha.com/input/?i=%5Csum(-1%2F3)%5En
$endgroup$
– dataphile
Mar 12 at 10:13
$begingroup$
sorry, I've added brackets now
$endgroup$
– dataphile
Mar 12 at 10:14
2
$begingroup$
OK, now you have a geometric series with first term $-1/m$ and common ratio $-1/m$, so its sum is $-1/(m+1)$, not $m/(m+1)$. Perhaps you want to start with $n=0$ instead of $n=1$?
$endgroup$
– Gerry Myerson
Mar 12 at 10:17
$begingroup$
Cool, thanks. I've fixed n.
$endgroup$
– dataphile
Mar 12 at 10:20
1
1
$begingroup$
Huh? Your series doesn't converge at all. The terms don't approach zero.
$endgroup$
– Gerry Myerson
Mar 12 at 10:11
$begingroup$
Huh? Your series doesn't converge at all. The terms don't approach zero.
$endgroup$
– Gerry Myerson
Mar 12 at 10:11
$begingroup$
See: wolframalpha.com/input/?i=%5Csum(-1%2Fm)%5En and wolframalpha.com/input/?i=%5Csum(-1%2F3)%5En
$endgroup$
– dataphile
Mar 12 at 10:13
$begingroup$
See: wolframalpha.com/input/?i=%5Csum(-1%2Fm)%5En and wolframalpha.com/input/?i=%5Csum(-1%2F3)%5En
$endgroup$
– dataphile
Mar 12 at 10:13
$begingroup$
sorry, I've added brackets now
$endgroup$
– dataphile
Mar 12 at 10:14
$begingroup$
sorry, I've added brackets now
$endgroup$
– dataphile
Mar 12 at 10:14
2
2
$begingroup$
OK, now you have a geometric series with first term $-1/m$ and common ratio $-1/m$, so its sum is $-1/(m+1)$, not $m/(m+1)$. Perhaps you want to start with $n=0$ instead of $n=1$?
$endgroup$
– Gerry Myerson
Mar 12 at 10:17
$begingroup$
OK, now you have a geometric series with first term $-1/m$ and common ratio $-1/m$, so its sum is $-1/(m+1)$, not $m/(m+1)$. Perhaps you want to start with $n=0$ instead of $n=1$?
$endgroup$
– Gerry Myerson
Mar 12 at 10:17
$begingroup$
Cool, thanks. I've fixed n.
$endgroup$
– dataphile
Mar 12 at 10:20
$begingroup$
Cool, thanks. I've fixed n.
$endgroup$
– dataphile
Mar 12 at 10:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The following series will converge at m:
$$sum_n=0^inftyfrac1.5m-2^n$$
answered Mar 12 at 17:54
dataphiledataphile
3791310
$endgroup$
add a comment |
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$begingroup$
The following series will converge at m:
$$sum_n=0^inftyfrac1.5m-2^n$$
answered Mar 12 at 17:54
dataphiledataphile
3791310
$endgroup$
add a comment |
$begingroup$
The following series will converge at m:
$$sum_n=0^inftyfrac1.5m-2^n$$
answered Mar 12 at 17:54
dataphiledataphile
3791310
$endgroup$
add a comment |
$begingroup$
The following series will converge at m:
$$sum_n=0^inftyfrac1.5m-2^n$$
answered Mar 12 at 17:54
dataphiledataphile
3791310
$endgroup$
The following series will converge at m:
$$sum_n=0^inftyfrac1.5m-2^n$$
answered Mar 12 at 17:54
dataphiledataphile
3791310
edited Mar 12 at 18:11
answered Mar 12 at 17:54
dataphiledataphile
3791310
answered Mar 12 at 17:54
dataphiledataphile
3791310
answered Mar 12 at 17:54
dataphiledataphile
3791310
3791310
add a comment |
add a comment |
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1
$begingroup$
Huh? Your series doesn't converge at all. The terms don't approach zero.
$endgroup$
– Gerry Myerson
Mar 12 at 10:11
$begingroup$
See: wolframalpha.com/input/?i=%5Csum(-1%2Fm)%5En and wolframalpha.com/input/?i=%5Csum(-1%2F3)%5En
$endgroup$
– dataphile
Mar 12 at 10:13
$begingroup$
sorry, I've added brackets now
$endgroup$
– dataphile
Mar 12 at 10:14
2
$begingroup$
OK, now you have a geometric series with first term $-1/m$ and common ratio $-1/m$, so its sum is $-1/(m+1)$, not $m/(m+1)$. Perhaps you want to start with $n=0$ instead of $n=1$?
$endgroup$
– Gerry Myerson
Mar 12 at 10:17
$begingroup$
Cool, thanks. I've fixed n.
$endgroup$
– dataphile
Mar 12 at 10:20