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Let $0 < \alpha < 1$. Can you choose $c > 0$ so that this modified harmonic series $$\sum_{k=1}^\infty \frac{1}{k^\alpha} \exp(-c k^{1-\alpha})$$ converges?

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Can't one show $e^{-ck^{\alpha}}<k^{-2}$ for $k$ sufficiently large? –  Gerry Myerson Oct 6 '12 at 12:14
    
It looks like it converges for any $c>0$. I tried different values of $\alpha$ and $c$ in Mathematica, they all converge. –  Patrick Li Oct 6 '12 at 13:02
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4 Answers

up vote 2 down vote accepted

$$\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ \alpha }{ e }^{ c{ k }^{ 1-\alpha } } } } =\quad \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ e^{ \alpha \ln { k } }{ e }^{ c{ k }^{ 1-\alpha } } } } =\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ e^{ \alpha \ln { k } +c{ k }^{ 1-\alpha } } } = } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ e^{ { k }^{ 1-\alpha }(c+\frac { \alpha \ln { k } }{ { k }^{ 1-\alpha } } ) } } }$$

Now let's make some observations to conclude:

General term is $a_k=\frac { 1 }{ e^{ { k }^{ 1-\alpha }(c+\frac { \alpha \ln { k } }{ { k }^{ 1-\alpha } } ) } } $ for which exists a $\eta $ such that for each n>$\eta$ we have that $a_k<\frac { 1 }{ { n }^{ 2 } }$. This proves that series converges.

If you want a proof of majoration consider that $\frac { \alpha \ln { k } }{ { k }^{ 1-\alpha } } $ tends to $0$ and that $ e^{ { k }^{ 1-\alpha }(c) }>{ n }^{ 2 }$ because (using log) $c{ k }^{ 1-\alpha}>2\ln { k }$.

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Hint: use the integral test and the fact that $$\frac d{dk} e^{-c\,k^{1-\alpha}}=(\alpha-1)c\frac {e^{-c\,k^{1 - \alpha}}}{k^\alpha}$$

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It converges for any $c>0$.

Consider that for arbitrary $x>0$, we have $e^{x}>x^{2}$.

So for any $c>0$, $e^{ck^{1-\alpha}}>c^{2}k^{2-2\alpha}$, then $0<\frac{1}{k^\alpha} \exp(-c k^{1-\alpha})<\frac{1}{c^{2}}\frac{1}{k^{2-\alpha}}$.

When $\alpha<1$, $\sum_{k=1}^{\infty}\frac{1}{k^{2-\alpha}}$ converges, so $\sum_{k=1}^\infty \frac{1}{k^\alpha} \exp(-c k^{1-\alpha})$ converges.

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Admittedly not the most direct route, but the idea is relatively simple:

Clearly for an arbitrary $k$ we have $0<\dfrac{1}{k^{\alpha}}\leq 1$ and thus $\dfrac{1}{k^{\alpha}e^{ck^{1-\alpha}}}\leq \dfrac{1}{e^{ck^{1-\alpha}}}\;\;\left(=e^{-ck^{1-\alpha}}\right)\;\;(*)$

Consider:

$$\int_1^{\infty} e^{-cx^{1-\alpha}}\;dx$$

Substitute $t$ for $c\cdot x^{1-\alpha}$:

$$\int_1^{\infty} e^{-cx^{1-\alpha}}\;dx=\frac{1}{c(1-\alpha)}\int_c^{\infty}t^{\frac{1}{1-\alpha}-1}e^{-t}\;dt$$

Recall the upper incomplete Gamma function:

$$\Gamma(a,x)=\int_x^{\infty} t^{a-1}e^{-t}\;dt$$

Therefore: $$\int_1^{\infty} e^{-cx^{1-\alpha}}\;dx=\frac{1}{c(1-\alpha)}\Gamma\left(\frac{1}{1-\alpha},c\right)$$

This is obviously finite for any $0<\alpha<1$, hence the series $\displaystyle\sum_{k=1}^{\infty} e^{-k^{1-\alpha}}$ must converge. From $(*)$:

$$\sum_{k=1}^{\infty}\dfrac{1}{k^{\alpha}e^{k^{1-\alpha}}}\leq \sum_{k=1}^{\infty} e^{-k^{1-\alpha}}$$

By comparison the original series converges also.

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