Cho \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^8}\)so sánh S với 1
So sánh S với \(\frac{1}{3}\)biết: \(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...........+\frac{2014}{5^{2014}}.\)
\(S=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}\)
\(5S=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}\)
\(\Rightarrow5S-S=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}\)
\(S=\frac{1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}}{4}\)
Xét \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(5A-A=1-\frac{1}{5^{2013}}\Leftrightarrow A=\frac{1-\frac{1}{5^{2013}}}{4}=\frac{1}{4}-\frac{1}{4.5^{2013}}\)
\(\Rightarrow S=\frac{1+\frac{1}{4}-\left(\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}\right)}{4}=\frac{5}{16}-\frac{\frac{1}{4.5^{2013}}+\frac{2014}{5^{2014}}}{4}< \frac{1}{3}\)
So sánh S với 2 biết :
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\)
ta có:S=\(\frac{481}{280}=1,717857143\)
=>S<2
vậy S<2
dạng 1 : so sánh
a) P = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}\)và Q = \(1\frac{3}{4}\)
dạng 2 : toán chứng minh
1. cho S = \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{130}\)chứng minh rằng : \(\frac{1}{4}< S< \frac{91}{330}\)
2. cho S = \(\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\). CMR : 3 < S < 8
3. CMR : \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{1999}}>1000\)
2.a) Vào question 126036
b) Vào question 68660
Cho S=\(\frac{1}{3\left(1+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}\)
So sánh S với \(\frac{3}{7}\)
So sánh \(S=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{9999}{10000}\)và \(\frac{1}{100}\)
So sánh \(S=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{9999}{10000}\)và \(\frac{1}{100}\)
Bài 1 : Cho \(S=\frac{1}{3\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{1}{97\left(\sqrt{48}+\sqrt{49}\right)}\)
So sánh S với \(\frac{3}{7}\)
\(tacó:...\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}>\frac{1}{3.2}=\frac{1}{\left(1+2.1\right).2.1}\)
\(\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}>\frac{1}{5.4}=\frac{1}{\left(1+2.2\right).2.2}\)
\(\frac{1}{7.\left(\sqrt{3}+\sqrt{4}\right)}>\frac{1}{7.6}=\frac{1}{\left(1+2..3\right).2.3}\)
....
\(\frac{1}{49.\left(\sqrt{48}+\sqrt{49}\right)}>\frac{1}{49.48}=\frac{1}{\left(1+2.48\right).2.48}\)
cộng vế theo vế ta đươc S =\(\frac{1}{\left(1+2.1\right).2}+\frac{1}{\left(1+2.2\right).2.2}+...+\frac{1}{\left(1+2.48\right).48.2}\)
\(=\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{10}+\frac{1}{21}+\frac{1}{36}+...+\frac{1}{4656}\right)\) < \(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4656}\right)\)
mà lại có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+..+\frac{1}{4656}\)
=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9312}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{96.97}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{97}=\frac{1}{2}-\frac{1}{97}=\frac{95}{194}\)
vậy S < \(\frac{95}{194}\)
mà \(\frac{95}{194}< \frac{3}{7}\)
=> S < \(\frac{3}{7}\)
KẾT LUẬN : S <\(\frac{3}{7}\)
Cho \(M=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}+\frac{7}{8}+\frac{8}{9}+\frac{9}{10}\)
So sánh M với 1
Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
Ta có:
1/2=0,5
2/3>0,6
<=>1/2+2/3>1,1>1
<=>1/2+2/3+3/4+...+9/10>1
Vì 1 = \(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)
\(\Rightarrow\)M > 1 vì \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};...;\frac{9}{10}>\frac{1}{10}\)
\(\Rightarrow M>1\)
So sánh tổng S=\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)với \(\frac{1}{2}\)
Ta có: 1/9 + 1/10 < 1/8+1/8 = 1/4
1/41+1/42< 1/40+1/40=1/20
=> 1/5+1/9+1/10+1/41+1/42<1/5+1/4+1/20=1/2
Vậy 1/5+1/9+1/10+1/41!+1/42<1/2