How to evaluate $\sum_{i=1}^n i^{2 i}$?

Question

Let, $$\mathcal{S}(n) = \sum_{i=1}^n i^{2 i}$$ for $n \in \mathbb{N}$

I will be completely honest. When I was returning from my physics tuition center, and suddenly this popped up in my head from nowhere. I don't know how.

However, I failed to evaluate the summation.

Please give me some hints how to handle the summation.

Answer 1

$$S_n=\sum_{i=1}^n i^{2 i}\qquad \implies\qquad S_n \sim n^{2n}$$

because the value of the sum is almost defined by the last term of the summation.

For example $S_{1000}\sim 1.00 \times 10^{6000}$ but $S_{1001}\sim 7.40 \times 10^{6006}$ that is to say almost ten millions times larger.

Answer 2

This is rather a long comment than an answer.

I am afraid that, according both to my long mathematical experience and already posted comments, some beautiful closed form for the sum, for instance, like that in Sophomore's_dream, can fail to exist.

On the other hand, sometimes ideas brightly popping in head indeed can be valuable. But usually they have to be verified later to assure the value. There is a cute and short related paper “Mathematical Creation” by Henri Poincaré.

Moreover, according to Wikipedia, Ramanujan, a great Indian number theorist, “credited his acumen to his family goddess, Namagiri Thayar (Goddess Mahalakshmi) of Namakkal. He looked to her for inspiration in his work and said he dreamed of blood drops that symbolised her consort, Narasimha. Later he had visions of scrolls of complex mathematical content unfolding before his eyes. He often said, “An equation for me has no meaning unless it expresses a thought of God””.

While asleep, I had an unusual experience. There was a red screen formed by flowing blood, as it were. I was observing it. Suddenly a hand began to write on the screen. I became all attention. That hand wrote a number of elliptic integrals. They stuck to my mind. As soon as I woke up, I committed them to writing.

Srinivasa Ramanujan

Answer 3

I can only say that $$\sum_{i=1}^n i^{2 i} \equiv \begin{cases} n/2 \pmod{4} \qquad \qquad \, \,\ \text{if $n$ even} \\ (n+1)/2 \pmod{4} \qquad \text{if $n$ odd} \end{cases}$$

I have no idea about finding this sum as we can easily see that how strange it behaved see fo $n=5$ your sum is $$1^2 +2^4+3^6+4^8+5^{10} = 9831907 $$

which is very close to $5^{2 \cdot 5}$ as @Claude Leibovici provided the approximation $n^{2n}$.

Answer 4

I played with a similar series a while ago. One can prove the following, which is essentially @Claude Leibovici's statement

$$ \sum_{i=1}^{n} i^i < (n+1)^{n+1} , n>=3 $$

Basically, check the statement for $n=2$ and $n=3$ and use induction.

Assume

$$ \sum_{i=1}^n i^i < (n+1)^{n+1} \text{ for } n \geq 2 \\ \sum_{i=1}^n i^i + (n+1)^{n+1} < (n+1)^{n+1} + (n+1)^{n+1} \\ \sum_{i=1}^n i^i + (n+1)^{n+1} < 2(n+1)^{n+1} < (n+1)(n+1)^{n+1} \text{for } n+1 \geq 2 \\ \sum_{i=1}^{n+1} i^i < (n+1)(n+1)^{n+1} = (n+1)^{n+2} < (n+2)^{n+2} $$

This proves it by induction.