So sánh: a) \(S = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + \frac{1}{{14}} + \frac{1}{{15}} + \frac{1}{{16}}

Câu hỏi số 389984:

Vận dụng

So sánh:

a) \(S = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + \frac{1}{{14}} + \frac{1}{{15}} + \frac{1}{{16}} + \frac{1}{{17}} + \frac{1}{{18}} + \frac{1}{{19}} + \frac{1}{{20}}\) và \(\frac{1}{2}\)

b) \(A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \ldots + \frac{1}{{{{100}^2}}}\) và \(\frac{3}{4}\)

c) \(B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{{2^{100}} - 1}}\) và \(100\)

d) \(D = \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + \ldots + \frac{1}{{100!}}\) và \(1\)

e) \(E = \frac{9}{{10!}} + \frac{{10}}{{11!}} + \frac{{11}}{{12!}} + \ldots + \frac{{999}}{{1000!}}\) và \(\frac{1}{{9!}}\)

f) \(F = \frac{{1 + 5 + {5^2} + \ldots + {5^9}}}{{1 + 5 + {5^2} + \ldots + {5^8}}}\) và \(G = \frac{{1 + 3 + {3^2} + \ldots + {3^9}}}{{1 + 3 + {3^2} + \ldots + {3^8}}}\)

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Câu hỏi:389984

Phương pháp giải

Để chứng minh ta giữ nguyên 1 phân số. Sau đó, các số hạng của tổng đó ta thay bằng phân số lớn hơn hoặc nhỏ hơn (kết hợp với các phương pháp nhóm hạng tử)

Áp dụng: \(n! = 1.2.3 \ldots \left( {n - 1} \right).n\)

Giải chi tiết

a) \(S = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + \frac{1}{{14}} + \frac{1}{{15}} + \frac{1}{{16}} + \frac{1}{{17}} + \frac{1}{{18}} + \frac{1}{{19}} + \frac{1}{{20}}\) và \(\frac{1}{2}.\)

Tổng \(S\) có \(10\) số hạng.

Vì \(\frac{1}{{11}} > \frac{1}{{10}};\,\,\frac{1}{{12}} > \frac{1}{{10}};\, \ldots ;\,\,\frac{1}{{20}} > \frac{1}{{10}}\) suy ra:

\(\begin{array}{l}S = \frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + \ldots + \frac{1}{{19}} + \frac{1}{{20}} > \frac{1}{{20}} + \frac{1}{{20}} + \ldots + \frac{1}{{20}}\\ \Rightarrow S > \frac{{1 + 1 + \ldots + 1}}{{20}}\\ \Rightarrow S > \frac{{1.10}}{{20}} = \frac{{10}}{{20}}\\ \Rightarrow S > \frac{1}{2}.\end{array}\)

Vậy \(S > \frac{1}{2} \cdot \)

b) \(A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \ldots + \frac{1}{{{{100}^2}}}\) và \(\frac{3}{4}.\)

Ta có: \(A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \ldots + \frac{1}{{{{100}^2}}}\)\( = \frac{1}{{2.2}} + \frac{1}{{3.3}} + \frac{1}{{4.4}} + \ldots + \frac{1}{{100.100}}\)

\(\left. \begin{array}{l}\frac{1}{{3.3}} < \frac{1}{{2.3}}\\\,\frac{1}{{4.4}} < \frac{1}{{3.4}}\\ \ldots \\\frac{1}{{100.100}} < \frac{1}{{99.100}}\end{array} \right\}\)\( \Rightarrow A = \frac{1}{{2.2}} + \frac{1}{{3.3}} + \frac{1}{{4.4}} + \ldots + \frac{1}{{100.100}} < \frac{1}{{2.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{99.100}}\)

Đặt \(B = \frac{1}{{2.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{99.100}}\). Ta có:

\(\begin{array}{l}B = \frac{1}{{2.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{99.100}}\\\,\,\,\,\, = \frac{1}{4} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{{99}} - \frac{1}{{100}}\\\,\,\,\,\, = \frac{1}{4} + \frac{1}{2} - \frac{1}{{100}}\\\,\,\,\,\, = \frac{1}{4} + \frac{2}{4} - \frac{1}{{100}}\\\,\,\,\,\, = \frac{3}{4} - \frac{1}{{100}}\\ \Rightarrow B < \frac{3}{4}\end{array}\)

Mà \(A < B\) suy ra \(A < \frac{3}{4}\).

Vậy \(A = \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \ldots + \frac{1}{{{{100}^2}}} < \frac{3}{4}\).

c) \(B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + .... + \frac{1}{{{2^{100}} - 1}}\) và \(100.\)

Để so sánh \(B\) với \(100\) ta chia \(B\) thành \(100\) nhóm.

\(\begin{array}{l}B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{{2^{100}} - 1}}\\\,\,\,\,\, = 1 + \frac{1}{{{2^1}}} + \frac{1}{3} + \frac{1}{{{2^2}}} + \frac{1}{5} + \ldots + \frac{1}{{{2^{100}} - 1}}\\\,\,\,\,\, = 1 + \left( {\frac{1}{{{2^1}}} + \frac{1}{3}} \right) + \left( {\frac{1}{{{2^2}}} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}} \right) + \left( {\frac{1}{{{2^3}}} + \ldots + \frac{1}{{15}}} \right) + \ldots + \left( {\frac{1}{{{2^{99}}}} + \ldots + \frac{1}{{{2^{100}} - 1}}} \right)\end{array}\)

Ta lại có:

\(\frac{1}{{{2^1}}} + \frac{1}{3} = \frac{1}{2} + \frac{1}{3} < \frac{1}{2} + \frac{1}{2} = \frac{1}{2} \cdot 2 = 1\)

\(\frac{1}{{{2^2}}} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} < \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{1}{4} \cdot 4 = 1\)

\( \ldots \)

\(\frac{1}{{{2^{99}}}} + \ldots + \frac{1}{{{2^{100}} - 1}} < \frac{1}{{{2^{99}}}} + \frac{1}{{{2^{99}}}} + \ldots + \frac{1}{{{2^{99}}}} = \frac{1}{{{2^{99}}}} \cdot {2^{99}} = 1\)

\(\begin{array}{l} \Rightarrow B = 1 + \left( {\frac{1}{{{2^1}}} + \frac{1}{3}} \right) + \left( {\frac{1}{{{2^2}}} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}} \right) + \left( {\frac{1}{{{2^3}}} + \ldots + \frac{1}{{15}}} \right) + \ldots + \left( {\frac{1}{{{2^{99}}}} + \ldots + \frac{1}{{{2^{100}} - 1}}} \right)\\ \Rightarrow B < 1 + 1 + 1 + \ldots + 1\end{array}\)

\( \Rightarrow B < 1.100 = 100\)

Vậy \(B = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{{2^{100}} - 1}} < 100.\)

d) \(D = \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + ..... + \frac{1}{{100!}}\) và \(1.\)

Ta có: \(\frac{1}{{2!}} = \frac{1}{{1.2}}\)

\(\begin{array}{l}\frac{1}{{3!}} = \frac{1}{{1.2.3}} = \frac{1}{{2.3}}\\\frac{1}{{4!}} = \frac{1}{{1.2.3.4}} < \frac{1}{{3.4}}\\ \ldots \end{array}\)

\(\frac{1}{{100!}} = \frac{1}{{1.2.3 \ldots 100}} < \frac{1}{{99.100}}\)

\( \Rightarrow \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{3!}} + \ldots + \frac{1}{{100!}} < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{99.100}}\)

\(\begin{array}{l} \Rightarrow \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{3!}} + \ldots + \frac{1}{{100!}} < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{{99}} - \frac{1}{{100}}\\ \Rightarrow D < 1 - \frac{1}{{100}} < 1\end{array}\)

Vậy \(\frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + \ldots + \frac{1}{{100!}} < 1\)

e) \(E = \frac{9}{{10!}} + \frac{{10}}{{11!}} + \frac{{11}}{{12!}} + ...... + \frac{{999}}{{1000!}}\) và \(\frac{1}{{9!}}.\)

Ta có:

\(\frac{9}{{10!}} = \frac{{10 - 1}}{{10!}} = \frac{{10}}{{10!}} - \frac{1}{{10!}} = \frac{1}{{9!}} - \frac{1}{{10!}}\)

\(\frac{{10}}{{11!}} = \frac{{11 - 1}}{{11!}} = \frac{{11}}{{11!}} - \frac{1}{{11!}} = \frac{1}{{10!}} - \frac{1}{{11!}}\)

\(\frac{{11}}{{12!}} = \frac{{12 - 1}}{{12!}} = \frac{{12}}{{12!}} - \frac{1}{{12!}} = \frac{1}{{11!}} - \frac{1}{{12!}}\)

\( \ldots \)

\(\frac{{999}}{{1000!}} = \frac{{1000 - 1}}{{1000!}} = \frac{{1000}}{{1000!}} - \frac{1}{{1000!}} = \frac{1}{{999!}} - \frac{1}{{1000!}}\)

\(\begin{array}{l} \Rightarrow \frac{9}{{10!}} + \frac{{10}}{{11!}} + \frac{{11}}{{12!}} + \ldots + \frac{{999}}{{1000!}} = \frac{1}{{9!}} - \frac{1}{{10!}} + \frac{1}{{10!}} - \frac{1}{{11!}} + \frac{1}{{11!}} - \frac{1}{{12!}} + \ldots + \frac{1}{{999!}} - \frac{1}{{1000!}}\\ \Rightarrow E = \frac{1}{{9!}} - \frac{1}{{1000!}} < \frac{1}{{9!}}\end{array}\)

\( \Rightarrow \frac{9}{{10!}} + \frac{{10}}{{11!}} + \frac{{11}}{{12!}} + \ldots + \frac{{999}}{{1000!}} < \frac{1}{{9!}}\)

Vậy \(\frac{9}{{10!}} + \frac{{10}}{{11!}} + \frac{{11}}{{12!}} + \ldots + \frac{{999}}{{1000!}} < \frac{1}{{9!}}\).

f) \(F = \frac{{1 + 5 + {5^2} + .... + {5^9}}}{{1 + 5 + {5^2} + ..... + {5^8}}}\) và \(G = \frac{{1 + 3 + {3^2} + .... + {3^9}}}{{1 + 3 + {3^2} + ..... + {3^8}}}.\)

Ta có:

\(\begin{array}{l}F = \frac{{1 + 5 + {5^2} + \ldots + {5^9}}}{{1 + 5 + {5^2} + \ldots + {5^8}}} = \frac{{1 + 5 + {5^2} + \ldots + {5^8} + {5^9}}}{{1 + 5 + {5^2} + \ldots + {5^8}}}\\\,\,\,\,\,\, = \frac{{1 + 5 + {5^2} + \ldots + {5^8}}}{{1 + 5 + {5^2} + \ldots + {5^8}}} + \frac{{{5^9}}}{{1 + 5 + {5^2} + \ldots + {5^8}}}\\\,\,\,\,\,\, = 1 + \frac{1}{{\frac{{1 + 5 + {5^2} + \ldots + {5^8}}}{{{5^9}}}}}\\\,\,\,\,\,\, = 1 + \frac{1}{{\frac{1}{{{5^9}}} + \frac{5}{{{5^9}}} + \frac{{{5^2}}}{{{5^9}}} + \ldots + \frac{{{5^8}}}{{{5^9}}}}}\\\,\,\,\,\,\,\, = 1 + \frac{1}{{\frac{1}{{{5^9}}} + \frac{1}{{{5^8}}} + \frac{1}{{{5^7}}} + \ldots + \frac{1}{5}}}\\G = \frac{{1 + 3 + {3^2} + \ldots + {3^9}}}{{1 + 3 + {3^2} + \ldots + {3^8}}} = \frac{{1 + 3 + {3^2} + \ldots + {3^8} + {3^9}}}{{1 + 3 + {3^2} + \ldots + {3^8}}}\\\,\,\,\,\,\, = \frac{{1 + 3 + {3^2} + \ldots + {3^8}}}{{1 + 3 + {3^2} + \ldots + {3^8}}} + \frac{{{3^9}}}{{1 + 3 + {3^2} + \ldots + {3^8}}}\\\,\,\,\,\,\, = 1 + \frac{1}{{\frac{{1 + 3 + {3^2} + \ldots + {3^8}}}{{{3^9}}}}}\\\,\,\,\,\,\, = 1 + \frac{1}{{\frac{1}{{{3^9}}} + \frac{3}{{{3^9}}} + \frac{{{3^2}}}{{{3^9}}} + \ldots + \frac{{{3^8}}}{{{3^9}}}}}\\\,\,\,\,\, = 1 + \frac{1}{{\frac{1}{{{3^9}}} + \frac{1}{{{3^8}}} + \frac{1}{{{3^7}}} + \ldots + \frac{1}{3}}}\end{array}\)

Bài toán đưa về so sánh \(\frac{1}{{{5^9}}} + \frac{1}{{{5^8}}} + \frac{1}{{{5^7}}} + \ldots + \frac{1}{5}\) và \(\frac{1}{{{3^9}}} + \frac{1}{{{3^8}}} + \frac{1}{{{3^7}}} + \ldots + \frac{1}{3}\).

Vì \(\frac{1}{3} > \frac{1}{5} \Rightarrow \frac{1}{{{3^2}}} > \frac{1}{{{5^2}}};\,\,\frac{1}{{{3^3}}} > \frac{1}{{{5^3}}}; \ldots ;\,\frac{1}{{{3^9}}} > \frac{1}{{{5^9}}}\)

\( \Rightarrow \frac{1}{{{3^9}}} + \frac{1}{{{3^8}}} + \frac{1}{{{3^7}}} + \ldots + \frac{1}{3} > \frac{1}{{{5^9}}} + \frac{1}{{{5^8}}} + \frac{1}{{{5^7}}} + \ldots + \frac{1}{5}\)

\( \Rightarrow \frac{1}{{\frac{1}{{{3^9}}} + \frac{1}{{{3^8}}} + \frac{1}{{{3^7}}} + \ldots + \frac{1}{3}}} < \frac{1}{{\frac{1}{{{5^9}}} + \frac{1}{{{5^8}}} + \frac{1}{{{5^7}}} + \ldots + \frac{1}{5}}}\)

\( \Rightarrow 1 + \frac{1}{{\frac{1}{{{3^9}}} + \frac{1}{{{3^8}}} + \frac{1}{{{3^7}}} + \ldots + \frac{1}{3}}} < 1 + \frac{1}{{\frac{1}{{{5^9}}} + \frac{1}{{{5^8}}} + \frac{1}{{{5^7}}} + \ldots + \frac{1}{5}}}\)

\( \Rightarrow G < F\) hay \(F < G\).

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