# Is it possible to put series in other series?

I've been working on a project for quite a long time but I found myself stuck at a step where I have to multiply elements of a series by elements of another series, which is dependent on the former one. Eventually, it should look like this: $$\sum_{i=1}^{\lfloor\frac{(n_f-11)}6\rfloor}(2i+3)\sum_{c=1}^i(2c-1)$$ where every result of the first series multiplies the various results of the second one, whose number increase with the value of i.

I would like to know if it is possible, as I haven't ever seen anything like this (I am little more than an amateur at Maths), and if it is not, I'd really appreciate some suggestions on alternative ways to do it.

• Is the last $-1$ part of the summand? – GFauxPas Jul 17 '16 at 18:00
• Yes, it is! The result of the first series (should) multiply each result of the second one $\sum_{c=1}^i2c-1$ – Estagon Jul 17 '16 at 18:37
• Then it should be in paretheses – GFauxPas Jul 17 '16 at 18:38
• Edited. Thanks for pointing that out! – Estagon Jul 17 '16 at 18:39

Notice that the inner sum is equal to

$$2 \sum \limits _{c=1} ^i c - \sum \limits _{c=1} ^i 1 = 2 \frac {i(i+1)} 2 - i = i^2 .$$

Your sum, then, becomes $$\sum \limits _{i=1} ^N (2i+3) i^2 = 2 \sum \limits _{i=1} ^N i^3 + 3 \sum \limits _{i=1} ^N i^2 = 2 \frac {N^2 (N+1)^2} 4 + 3 \frac {N(N+1)(2N+1)} 6 = \frac {N^4 + 4N^3 + 4N^2 + N} 2 ,$$

where $N = \lfloor \dfrac {n_f-11} 6 \rfloor$.

The sums of powers that I have used are well known, you can find a brief list on Wikipedia.

Hint. One may recall that $$\sum_{c=1}^i(2c-1)=i^2$$ and that $$\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6, \quad \quad \sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4.$$

Thus your initial sum rewrites $$\sum_{i=1}^{\lfloor\frac{(n_f-11)}6\rfloor}(2i+3)\sum_{c=1}^i(2c-1)=\sum_{i=1}^{\lfloor\frac{(n_f-11)}6\rfloor}(2i+3)\cdot i^2=2\sum_{i=1}^{n}i^3+3\sum_{i=1}^{n}i^2$$ with $\displaystyle n=\lfloor\frac{(n_f-11)}6\rfloor$.

• @Estagon Did my above hint help you? Thanks. – Olivier Oloa Jul 17 '16 at 18:51