Finite sum of $1/n^{delta}$
0
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How does one show
$sum_{n leq x/M} frac{1}{n^{delta}} approx frac{(N/M)^{delta}}{1-delta}$, where $n leq N$ and $delta>0$.
I assume this has something to do with Taylor series, but im not sure around what number.
It also might be useful to use the Euler-Mac Laurin summation formula...
summation taylor-expansion
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
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add a comment |
0
$begingroup$
How does one show
$sum_{n leq x/M} frac{1}{n^{delta}} approx frac{(N/M)^{delta}}{1-delta}$, where $n leq N$ and $delta>0$.
I assume this has something to do with Taylor series, but im not sure around what number.
It also might be useful to use the Euler-Mac Laurin summation formula...
summation taylor-expansion
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
$endgroup$
$begingroup$
Is your sum from $n=1$ to $frac xM$? If so, shouldn't the sum depend on $x$? Or does $N=frac xM$? Please clarify. It looks like it comes from turning the sum into an integral.
$endgroup$
– Ross Millikan
Dec 10 '18 at 16:51
$begingroup$
These are the generalized harmonic numbers. I suspect that they wouldn't grow like any power of $x/M$.
$endgroup$
– Will Fisher
Dec 10 '18 at 16:56
$begingroup$
Do you also want $delta<1?$ Otherwise the right-sand side would be negative.
$endgroup$
– saulspatz
Dec 10 '18 at 16:58
add a comment |
0
0
0
$begingroup$
How does one show
$sum_{n leq x/M} frac{1}{n^{delta}} approx frac{(N/M)^{delta}}{1-delta}$, where $n leq N$ and $delta>0$.
I assume this has something to do with Taylor series, but im not sure around what number.
It also might be useful to use the Euler-Mac Laurin summation formula...
summation taylor-expansion
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
$endgroup$
How does one show
$sum_{n leq x/M} frac{1}{n^{delta}} approx frac{(N/M)^{delta}}{1-delta}$, where $n leq N$ and $delta>0$.
I assume this has something to do with Taylor series, but im not sure around what number.
It also might be useful to use the Euler-Mac Laurin summation formula...
summation taylor-expansion
summation taylor-expansion
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
asked Dec 10 '18 at 16:43
usere5225321usere5225321
622412
622412
$begingroup$
Is your sum from $n=1$ to $frac xM$? If so, shouldn't the sum depend on $x$? Or does $N=frac xM$? Please clarify. It looks like it comes from turning the sum into an integral.
$endgroup$
– Ross Millikan
Dec 10 '18 at 16:51
$begingroup$
These are the generalized harmonic numbers. I suspect that they wouldn't grow like any power of $x/M$.
$endgroup$
– Will Fisher
Dec 10 '18 at 16:56
$begingroup$
Do you also want $delta<1?$ Otherwise the right-sand side would be negative.
$endgroup$
– saulspatz
Dec 10 '18 at 16:58
add a comment |
$begingroup$
Is your sum from $n=1$ to $frac xM$? If so, shouldn't the sum depend on $x$? Or does $N=frac xM$? Please clarify. It looks like it comes from turning the sum into an integral.
$endgroup$
– Ross Millikan
Dec 10 '18 at 16:51
$begingroup$
These are the generalized harmonic numbers. I suspect that they wouldn't grow like any power of $x/M$.
$endgroup$
– Will Fisher
Dec 10 '18 at 16:56
$begingroup$
Do you also want $delta<1?$ Otherwise the right-sand side would be negative.
$endgroup$
– saulspatz
Dec 10 '18 at 16:58
$begingroup$
Is your sum from $n=1$ to $frac xM$? If so, shouldn't the sum depend on $x$? Or does $N=frac xM$? Please clarify. It looks like it comes from turning the sum into an integral.
$endgroup$
– Ross Millikan
Dec 10 '18 at 16:51
$begingroup$
Is your sum from $n=1$ to $frac xM$? If so, shouldn't the sum depend on $x$? Or does $N=frac xM$? Please clarify. It looks like it comes from turning the sum into an integral.
$endgroup$
– Ross Millikan
Dec 10 '18 at 16:51
$begingroup$
These are the generalized harmonic numbers. I suspect that they wouldn't grow like any power of $x/M$.
$endgroup$
– Will Fisher
Dec 10 '18 at 16:56
$begingroup$
These are the generalized harmonic numbers. I suspect that they wouldn't grow like any power of $x/M$.
$endgroup$
– Will Fisher
Dec 10 '18 at 16:56
$begingroup$
Do you also want $delta<1?$ Otherwise the right-sand side would be negative.
$endgroup$
– saulspatz
Dec 10 '18 at 16:58
$begingroup$
Do you also want $delta<1?$ Otherwise the right-sand side would be negative.
$endgroup$
– saulspatz
Dec 10 '18 at 16:58
add a comment |
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$begingroup$
Is your sum from $n=1$ to $frac xM$? If so, shouldn't the sum depend on $x$? Or does $N=frac xM$? Please clarify. It looks like it comes from turning the sum into an integral.
$endgroup$
– Ross Millikan
Dec 10 '18 at 16:51
$begingroup$
These are the generalized harmonic numbers. I suspect that they wouldn't grow like any power of $x/M$.
$endgroup$
– Will Fisher
Dec 10 '18 at 16:56
$begingroup$
Do you also want $delta<1?$ Otherwise the right-sand side would be negative.
$endgroup$
– saulspatz
Dec 10 '18 at 16:58