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i have to prove that the sequence {a

[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?

_{k}} 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?

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