Question

[Nguyễn Mạnh Hùng]

1/2^2+1/3^2+1/4^2+...+1/20^2. chứng tỏ rằng A<1

Answer 1

[Nguyễn Việt Hoàn]

We want to evaluate the sum \(A = \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \hdots + \frac{1}{2 0^{2}}\), and show that \(A < 1\).

Step-by-step explanation:

1. Express the sum mathematically:

\(A = \sum_{n = 2}^{20} \frac{1}{n^{2}} .\)

1. Estimate the upper bound of the series:
   The term \(\frac{1}{n^{2}}\) decreases as \(n\) increases, so the sum converges. To prove \(A < 1\), note that the largest contributions to the sum come from the first few terms.
2. Compute the approximate sum for small \(n\):
3. • For \(n = 2\): \(\frac{1}{2^{2}} = \frac{1}{4} = 0.25\),
   • For \(n = 3\): \(\frac{1}{3^{2}} = \frac{1}{9} \approx 0.111\),
   • For \(n = 4\): \(\frac{1}{4^{2}} = \frac{1}{16} = 0.0625\),
   • Continue this pattern for \(n = 5 , 6 , \ldots , 20\).
4. Sum upper bound comparison:
   Since \(\frac{1}{n^{2}} < \frac{1}{n \left(\right. n - 1 \left.\right)}\) for \(n \geq 2\), the sum \(\sum_{n = 2}^{20} \frac{1}{n^{2}}\) is smaller than a telescoping series that can be computed easily. Comparing directly, adding all computed terms does **converging key simpl hint

Answer 2

[Kiều Vũ Linh]

Ta có:

...

Cộng vế với vế, ta được:

Từ (1) và (2) suy ra: