Question

Given two series S₁ = 1 - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ +.....+ $\frac{1}{2019}$ and S₂ = $\frac{1}{1010}$ + $\frac{1}{1011}$ + $\frac{1}{1012}$ +.....+ $\frac{1}{2019}$, then

• A. S₁ < S₂
• B. S₁ > S₂
• C. S₁ = S₂
• D. S₁+S₂ < 2

Answer 1

Answer: (C), (D)

C

Answer 2

The series S₁ can be written as: $S₁ = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{2019}$

We can rewrite S₁ by grouping terms: $S₁ = (1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2019}) - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots + \frac{1}{2018})$

To relate this to the harmonic series ($H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$), we can add and subtract the even terms: $S₁ = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2019}) - 2(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2018})$ $S₁ = H_{2019} - 2 \cdot \frac{1}{2} (1 + \frac{1}{2} + \dots + \frac{1}{1009})$ $S₁ = H_{2019} - H_{1009}$

Now, expanding $H_{2019} - H_{1009}$: $H_{2019} = (1 + \frac{1}{2} + \dots + \frac{1}{1009}) + (\frac{1}{1010} + \frac{1}{1011} + \dots + \frac{1}{2019})$ $H_{1009} = 1 + \frac{1}{2} + \dots + \frac{1}{1009}$ So, $H_{2019} - H_{1009} = (\frac{1}{1010} + \frac{1}{1011} + \dots + \frac{1}{2019})$.

This is precisely the series S₂. Thus, $S₁ = S₂$.

Now consider $S₁ + S₂ < 2$. Since $S₁ = S₂$, this inequality becomes $2S₁ < 2$, or $S₁ < 1$. Let's check if $S₁ < 1$: $S₁ = 1 - (\frac{1}{2} - \frac{1}{3}) - (\frac{1}{4} - \frac{1}{5}) - \dots - (\frac{1}{2018} - \frac{1}{2019})$ Each term in parentheses is positive, e.g., $(\frac{1}{2} - \frac{1}{3}) = \frac{1}{6}$. Therefore, $S₁ = 1 - (\text{a sum of positive terms})$, which implies $S₁ < 1$. Consequently, $S₁ + S₂ < 2$ is also true.

Both options (C) and (D) are correct. However, option (C) establishes the direct equality between the two series, which is the primary relationship derived.