Quick formula for approximate sigma sum

Question

Is there a quick way to get an approximate answer to this sum. My maths is a little out of practice:

$$ M(1+i/360)^{360}\left(1+\sum_{k=1}^{11}(1+i/360)^{-30k}\right) $$

This is a formula for compound interest. Say i=0.02, M=1000

Answer 1

Consider the summation $$S_n=\sum_{k=1}^{n}\left(1+\frac i{360}\right)^{-30k}=\sum_{k=1}^{n}\frac{1}{\left(\left(1+\frac{i}{360}\right)^{30}\right)^k}$$

Now, using Taylor $$\frac 1{\left(1+\frac{i}{360}\right)^{30}}=1-\frac{i}{12}+\frac{31 i^2}{8640}-\frac{31 i^3}{291600}+O\left(i^4\right)$$ $$\frac{1}{\left(\left(1+\frac{i}{360}\right)^{30}\right)^k}=1-\frac{i }{12}k+\frac{i^2 }{8640}k (30 k+1)-\frac{i^3 }{4665600}k (15 k+1) (30 k+1)+O\left(i^4\right)$$

Now expand the trems with $k$ and use Faulhaber's formulae and you will arrive to $$S_n=n-\frac{1}{24} i (n (n+1))+\frac{i^2 n (n+1) (20 n+11)}{17280}-\frac{i^3 (n (n+1) (15 n+1) (15 n+16))}{9331200}+O\left(i^{4}\right)$$

Using it for $i=\frac 2{100}$ and $n=11$ the approximation will be $$S_{11}=\frac{529288138247}{48600000000}\approx 10.89070243$$ while the "exact" result would be $10.89070245$

Answer 2

So I think I can resolve this using approximation by definite integrals

https://en.wikipedia.org/wiki/Summation?fbclid=IwAR1hebwyinD2rfAWnXrZl679eRM9GBKA5BehD0xfiCaEMXDn-W0AxQvW-b0#Approximation_by_definite_integrals

$$ M(1+i/360)^{360} (1+ \int_{x=1}^{12}{(1.00005555)^{-30x}} dx )$$

Answer 3

You may use $(1+\frac{i}{360})^{360} \approx \exp(i)$.

The summation, you can write as

$$\sum_{i=1}^{11}\left(1+\frac{i}{360}\right)^{-30k}$$

$$\sum_{i=1}^{11}\left(\frac{1}{\left(1+\frac{i}{360}\right)^{30}}\right)^{k}$$

Using the binomial expansion (ignoring the higher order terms),

$$\sum_{i=1}^{11}\left(\frac{1}{\left(1+30\frac{i}{360}\right)}\right)^{k}$$

$$\sum_{i=1}^{11}\left(\frac{1}{1+\frac{i}{12}}\right)^{k}$$

Further

$$\sum_{i=1}^{11}\left(\left(1+\frac{i}{12}\right)^{-1}\right)^{k}$$

Upon further expansion and ignoring the higher order terms

$$\sum_{i=1}^{11}\left(1-k\frac{i}{12}\right)$$ i.e. $$11\left(1-\frac{i}{2}\right)$$

The expression approximates to $$M\exp(i)\left[1+11\left(1-\frac{i}{2}\right)\right]$$