I am looking for the tightest known bound for the sum  

$$\sum\_{\substack{1\leq k\leq j^\alpha \\ k\mid j}}k^\lambda$$
where $j$ is a large positive integer, $\alpha\in(0,1)$ and $\lambda\geq 1$. 


I am also interested whether it is possible to give any better estimate on the sum  

$$\sum\_{j=n}^{n^2} \frac{1}{j^\lambda} \sum\_{\substack{1\leq k\leq j^\alpha \\ k\mid j}}k^\lambda$$
with the same parameters $\alpha, \lambda$ and for some large positive integer $n$ (if it makes any difference one can think on $n$ as being a large prime number), rather than just using the bound for the previous sum.



Sum of small divisors with powers
The second sum, can be rewritted as
$$\sum\_{k=1}^{n^{2\alpha}} k^\lambda \sum\_{j=\max\{n,k^{1/\alpha}\}\atop k\mid j}^{n^2} \frac{1}{j^\lambda} = \sum\_{k=1}^{n^{2\alpha}} k^\lambda \sum\_{\ell=\lceil \max\{n/k,k^{1/\alpha-1}\}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{(k\ell)^\lambda} = \sum\_{k=1}^{n^{2\alpha}} \sum\_{\ell=\lceil \max\{n/k,k^{1/\alpha-1}\}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda}$$
$$=\sum\_{k=1}^{n^{\alpha}}\sum\_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} + \sum\_{k=n^{\alpha}}^{n^{2\alpha}} \sum\_{\ell=\lceil k^{1/\alpha-1}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda}.$$
The inner sums can be bounded by the integrals:
$$\int\_{L+1}^{n^2/k} \frac{d\,t}{t^\lambda} \leq \sum\_{\ell=\lceil L\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq \int\_{L-1}^{n^2/k} \frac{d\,t}{t^\lambda}.$$
E.g., for $\lambda>1$, we have
$$C\_1k^{\lambda-1} \leq \sum\_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq C\_2k^{\lambda-1},$$
where 
$$C\_1=\frac{n^{\lambda-1}-1}{(\lambda-1)n^{2(\lambda-1)}}, \qquad C\_2= \frac{(n-n^\alpha)^{1-\lambda}-1}{(\lambda-1)n^{2(\lambda-1)}}.$$
Then 
$$\sum\_{k=1}^{n^{\alpha}} \sum\_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \geq \sum\_{k=1}^{n^{\alpha}} C\_1 k^{\lambda-1}\geq \frac{C\_1}{\lambda} n^{\alpha\lambda}$$
and
$$\sum\_{k=1}^{n^{\alpha}} \sum\_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq \sum\_{k=1}^{n^{\alpha}} C\_2 k^{\lambda-1}\leq \frac{C\_2}{\lambda} (n^\alpha+1)^{\lambda}.$$
The other sum is estimated similarly.


