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Find ${ \alpha }$ such that ${ \sum_{k=1}^{ \infty } \left(k \alpha - k[ \alpha ]\right) }$ is covergence.

(${[x]}$ is is the largest integer not greater than $x$. ${ \alpha }$ is a positive real number. )

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If this is a homework question, then please add the tag (homework). Also, this question showed no attempt or effort. –  user2468 Jul 28 '12 at 2:32
    
If this problem change "Find ${ \alpha }$ such that ${ \sum_{k=1}^{ \infty } \left(k \alpha - [k \alpha ]\right) }$ is covergence. (${[x]}$ is is the largest integer not greater than $x$. ${ \alpha }$ is a positive real number. )" –  Go Jiyoung Jul 28 '12 at 2:58

3 Answers 3

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This is an answer to the second version of the question, viz. find for which $\alpha > 0$, $ \sum_{k=1}^\infty (k\alpha-\lfloor k\alpha \rfloor) $ converges.

When $\alpha$ is an integer, the series converges to zero since each term is zero.

When $\alpha = C + a/b$ is rational ($C$ an integer, $0 < a < b$), for each $k = nb + 1$ we have $$ k\alpha - \lfloor k\alpha \rfloor = (nb+1) C + na + a/b - \lfloor (nb+1) C + na + a/b \rfloor = a/b, $$ and so the series diverges.

When $\alpha > 0$ is irrational, the term $n\alpha - \lfloor n\alpha \rfloor$ is "asymptotically uniform" over $[0,1]$ (see equidistribution theorem). In particular, there are infinitely many terms which are at least $1/2$, and so again the series diverges.

Summarizing, the series converges if and only if $\alpha$ is integral.

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Hint: The Term Test must be satisfied,

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If $\alpha$ is an integer, then $k\alpha -k\lfloor \alpha\rfloor=0$ for every positive integer $k$, so our series converges.

If $\alpha$ is not an integer, then $\sum_{k=1}^\infty \left(k\alpha -k\lfloor \alpha\rfloor\right)$ does not converge.

For let $\alpha=m+t$, where $m$ is an integer and $0\lt t\lt 1$. Then $$k\alpha -k\lfloor \alpha\rfloor=k(m+t)-km=kt.$$ But $\sum_{k=1}^n kt$ grows without bound as $n\to\infty$.

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