Is there a way to rewrite this expression as a single sum?

$$\sum _{n=1}^k \frac{3^{-2 n}}{n}-\sum _{n=1}^{2 k} \frac{2\ 3^{-n}}{n}$$

It can be rewritten as

$$\sum _{n=1}^k \frac{3^{-2 n}-2\ 3^{-n}}{n}-\sum _{n=k+1}^{2 k} \frac{2\ 3^{-n}}{n}$$

but that did not help me to simplify it any further.

If the upper limits were the same in both sums then it would be trivial but the second limit is twice the first one.


2 Answers 2


The substitution $j=2n$ gives $$\sum _{n=1}^k \frac{3^{-2 n}}{n} = \sum _{j=2,\ j \text{ even}}^{2k} \frac{2 \times 3^{-j}}{j}$$

so \begin{align*}\sum _{n=1}^k \frac{3^{-2 n}}{n}-\sum _{n=1}^{2 k} \frac{2\ 3^{-n}}{n} &= \sum _{j=2,\ j \text{ even}}^{2k} \frac{2 \times 3^{-j}}{j} -\sum _{n=1}^{2 k} \frac{2\times 3^{-n}}{n} =-\sum _{j=1,\ j \text{ odd}}^{2k} \frac{2 \times 3^{-j}}{j} \end{align*}

i.e. finally $$\boxed{\sum _{n=1}^k \frac{3^{-2 n}}{n}-\sum _{n=1}^{2 k} \frac{2\ 3^{-n}}{n} = \sum _{n=1}^{k} \frac{2 \times 3^{1-2n}}{1-2n}}$$

  • 1
    $\begingroup$ I just found one formula too, look at my answer, but yours is better. $\endgroup$ Dec 3, 2023 at 23:06

I just found this:

$$\boxed{\sum _{n=1}^{2 k} \frac{3^{-n} \left((-1)^n-1\right)}{n}}$$

$$\sum _{n=1}^k \frac{3^{-2 n}}{n}-\sum _{n=1}^{2 k} \frac{2\ 3^{-n}}{n}=\sum _{n=1}^{2 k} \frac{3^{-2 n/2} \left((-1)^n+1\right)/2}{n/2}-\sum _{n=1}^{2 k} \frac{2\ 3^{-n}}{n}=\sum _{n=1}^{2 k} \left(\frac{3^{-n} \left((-1)^n+1\right)}{n}-\frac{2\ 3^{-n}}{n}\right)=\sum _{n=1}^{2 k} \frac{3^{-n} \left((-1)^n+1\right)-2\ 3^{-n}}{n}=\boxed{\sum _{n=1}^{2 k} \frac{3^{-n} \left((-1)^n-1\right)}{n}}$$

$$\sum _{n=1}^{2 k} \frac{3^{-n} \left((-1)^n-1\right)}{n}=\sum _{n=1}^k \left(\frac{3^{-(2 n-1)} \left((-1)^{2 n-1}-1\right)}{2 n-1}+\frac{3^{-(2 n)} \left((-1)^{2 n}-1\right)}{2 n}\right)=\sum _{n=1}^k \left(\frac{3^{-(2 n-1)} (-2)}{2 n-1}+\frac{3^{-(2 n)} (0)}{2 n}\right)=\boxed{-2 \sum _{n=1}^k \frac{3^{1-2 n}}{2 n-1}}$$


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