How do I write the following series in summation notation; 4 + 1/5 + .3 + 1/(3 + sqrt 2) + 1/(9 + sqrt 3) + 1/(27 + sqrt 4) +.....?

Question

I have tried a wide variety of exponential notation, logirithms, etc.  Any help to get moving in the right direction would be helpful

Answer 1

Part of the answer is ∑1/(3^k+√(k+1)) for k=1 to n. If we put k=0, we get 1/(1+1)=1/2 or 0.5. This does not fit in with the first terms. It seems only to apply when k>0.

Take the general term of the series. The fraction can be rationalised: 1/(3^k+√(k+1) * (3^k-√(k+1))/(3^k-√(k+1)). This rationalises to (3^k-√(k+1))/(3^2k-k-1). For k=3 this is  (27-2)/(729-4)=25/725=1/29=1/(27+2). For k=1, rationalisation produces (3-√2)/(9-2)=(3-√2)/7. Note that the denominator is always rational. The first three terms 4+1/5+0.3=4.5. As it happens this can be split into 4 +1/2, where 1/2 is the original summation starting with k=0. The series then becomes:

4+∑1/(3^k+√(k+1)) for k=0 to n; or 4+∑(3^k-√(k+1))/(3^2k-k-1) in rational form (use other form when k=0 to avoid division by zero).