0
$\begingroup$

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?

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

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

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) $c{ k }^{ 1-\alpha}>2\ln { k }$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.