May 29, 18 · Transcript Ex 41, 9 Prove the following by using the principle of mathematical induction for all n ∈ N 1/2 1/4 1/8 1/2𝑛 = 1 – 1/2𝑛 Let P(n) 1/2 1/4 1/8 1/2𝑛 = 1 – 1/2𝑛 For n = 1, we have LHS = 1/2 RHS = 1 – 1/21 = 1/2 Hence, LHS = RHS , ∴ P(n) is true for n = 1 Assume P(k) is true 1/2 1/4 1/8 1/2𝑘 = 1 – 1/2𝑘 WeYou can put this solution on YOUR website!$$\ln(n1)\le\sum_{i=1}^n\frac1i\le\ln(n)1$$ This is a rather tight upper limit and lower limit you can use to approximate your answer One could also note that $$\sum_{i=1}^n\frac1i=\int_0^1\sum_{i=0}^{n1}x^i\ dx=\int_0^1\frac{1x^n}{1x}\ dx$$ We also have the EulerMaclaurin expansion C Exercises Display The N Terms Of Harmonic Series And Their Sum W3resource 1-1/2+1/3-...+1/n formula