The discussion focuses on evaluating the sum $$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}$$. Participants express appreciation for Sudharaka's solution, indicating its clarity and effectiveness. The evaluation involves understanding the behavior of the terms as n increases, particularly the influence of the square root term on the denominator. This sum is relevant in mathematical series and convergence analysis.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in series evaluation and convergence analysis.
lfdahl said:Evaluate the sum:$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}$$
Sudharaka said:$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3+\sqrt{3^{2017}}}+\frac{1}{3^2+\sqrt{3^{2017}}}+\cdots+\frac{1}{3^{2015}+\sqrt{3^{2017}}}+\frac{1}{3^{2016}+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}$$
Observe that taking the first and last terms;
$$\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{\sqrt{3^{2017}}(1+\sqrt{3^{2017}})}=\frac{1}{\sqrt{3^{2017}}}$$
Generally this holds for all $i$ and $n-i$ terms taken in pairs. Thus our sum reduces to,
$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\left(\frac{2018}{2}\right)\frac{1}{\sqrt{3^{2017}}}=\frac{1009}{\sqrt{3^{2017}}}$$