Determine the integer part of $\log_2{2011}+\log_{3}2012+\log_{4}2013+\log_{5}2014+\log_{6}2015+\log_{7}2016$
Solution.
- $\log_2 2=1$
- $3^{12}=531441>524288=2^{19}$ and so $\log_23>\frac{19}{12}$
- $\log_24=2$
- $5^4=625>512=2^9$ and so $\log_25>\frac{9}{4}$
- $\log_23>\frac{19}{12}$ and so $\log_26>\frac{31}{12}$
- $7^5=16807>16384=2^{14}$ and so $\log_27>\frac{14}5$
So $\sum_{k=2}^7\frac 1{\log_2k}<1+\frac{12}{19}$ $+\frac 12+\frac 49+\frac{12}{31}+\frac 5{14}$ $=\frac{123205}{37107}$
So $\sum_{k=2}^7\frac{\log_2(2009+k)}{\log_2k}<\log_22048\sum_{k=2}^7\frac 1{\log_2k}$ $<11\frac{123205}{37107}$ $<37$
- $\log_2 2=1$
- $3^{5}=243<256=2^{8}$ and so $\log_23<\frac 85$
- $\log_24=2$
- $5^3=125<128=2^7$ and so $\log_25<\frac 73$
- $\log_23<\frac 85$ and so $\log_26<\frac{13}{5}$
- $7^6=117649<131072=2^{17}$ and so $\log_27<\frac{17}6$
So $\sum_{k=2}^7\frac 1{\log_2k}>1+\frac{5}{8}$ $+\frac 12+\frac 37+\frac{5}{13}+\frac 6{17}$ $=\frac{40731}{12376}$
Note also that $\left(1+\frac{37}{2011}\right)^{20}<e^{\frac{20\times 37}{2011}}$ $<3^{\frac {740}{2011}}$ $<3^{\frac 12}<2$ So $2^{\frac 1{20}}>1+\frac{37}{2011}=\frac{2048}{2011}$
So $2011>2^{11-\frac 1{20}}$ and so $\log_22011>\frac{219}{20}$
So $\sum_{k=2}^7\frac{\log_2(2009+k)}{\log_2k}>\log_22011\sum_{k=2}^7\frac 1{\log_2k}$ $>\frac{219}{20}\frac{40731}{12376}$ $>36$
Hence the requested integer part $\boxed {36}$
Question: Do you have any simpler method to solve it?