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OpenStudy (anonymous):
Sum of n terms in a harmonic progression
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OpenStudy (anonymous):
I am asking
OpenStudy (ikram002p):
do you have a harmonic progression
?
OpenStudy (anonymous):
no in general
OpenStudy (ikram002p):
but he ask about Sum of n terms , so integrate might work :D
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OpenStudy (ikram002p):
or that would be approximation ?
ganeshie8 (ganeshie8):
how does one integrate a floor function
\[\large \int \limits_1^{\infty}\dfrac{1}{[x] } - \dfrac{1}{x} dx\]
ganeshie8 (ganeshie8):
*\[\large \int \limits_1^{\infty}\dfrac{1}{\lfloor x \rfloor } - \dfrac{1}{x} dx\]
OpenStudy (ikram002p):
by common sense mmm
its descrete so try to make conjecture
OpenStudy (ikram002p):
but i think we dnt need floor function mmm
1/x it self should be fine approximation
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ganeshie8 (ganeshie8):
\[\int \limits_1^n \lfloor x\rfloor = 1 + 2 + 3 + \cdots (n-1)\]
?
ganeshie8 (ganeshie8):
\[\int \limits_1^{n} \dfrac{1}{\lfloor x\rfloor} dx = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n-1}\]
ganeshie8 (ganeshie8):
hmm
ganeshie8 (ganeshie8):
\[\ln n \ne \int \limits_1^{n} \dfrac{1}{\lfloor x\rfloor} dx = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n-1} \]
OpenStudy (ikram002p):
lets refare to integration definition
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OpenStudy (ikram002p):
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