- #1
murshid_islam
- 458
- 19
i have to prove that the sequence {ak} is decreasing, where
[tex]\{a_k\} = \{\left({1+\frac{1}{k}}\right)^{-k}\}[/tex]
this is what i did:
[tex]a_k = {\left(\frac{k}{k+1}}\right)^k[/tex]
[tex]a_{k+1}-a_{k}[/tex]
[tex]= {\left(\frac{k+1}{k+2}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)}^{k}[/tex]
[tex]= {\left(\frac{1+\frac{1}{k}}{1+\frac{2}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]
[tex]< {\left(\frac{1+\frac{1}{k}}{1+\frac{1}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]
[tex]= 1 - \left({\frac{k}{k+1}}\right)^{k}[/tex]
[tex]< 1-1[/tex] since [tex]\left({\frac{k}{k+1}}\right)^{k} < 1[/tex]
therefore, [tex]a_{k+1}-a_{k} < 0[/tex]
therefore, the sequence is decreasing.
am i right?
[tex]\{a_k\} = \{\left({1+\frac{1}{k}}\right)^{-k}\}[/tex]
this is what i did:
[tex]a_k = {\left(\frac{k}{k+1}}\right)^k[/tex]
[tex]a_{k+1}-a_{k}[/tex]
[tex]= {\left(\frac{k+1}{k+2}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)}^{k}[/tex]
[tex]= {\left(\frac{1+\frac{1}{k}}{1+\frac{2}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]
[tex]< {\left(\frac{1+\frac{1}{k}}{1+\frac{1}{k}}}\right)^{k+1}-{\left(\frac{k}{k+1}}\right)^{k}[/tex]
[tex]= 1 - \left({\frac{k}{k+1}}\right)^{k}[/tex]
[tex]< 1-1[/tex] since [tex]\left({\frac{k}{k+1}}\right)^{k} < 1[/tex]
therefore, [tex]a_{k+1}-a_{k} < 0[/tex]
therefore, the sequence is decreasing.
am i right?
Last edited: