Prove that $ dfrac{1}{5}+ dfrac{1}{9}+ dfrac{1}{13}+ldots+ dfrac{ 1}{2005}< dfrac{ 7}{4}$.












2














Prove that
$ dfrac{1}{5}+ dfrac{1}{9}+ dfrac{1}{13}+ldots+ dfrac{ 1}{2005}< dfrac{ 7}{4}$.



Solution
$sum_{k=1}^{501}frac 1{4k+1}$ $=frac 15+sum_{k=2}^{501}frac 1{4k+1}$ $<frac 15+int_1^{501}frac{dx}{4x+1}$ $=frac 15+left[frac{ln(4x+1)}4right]_1^{501}$



So $sum_{k=1}^{501}frac 1{4k+1}$ $<frac 15+frac{ln 2005-ln 5}4$ $=frac 15+frac{ln 401}4$ $<frac 74$
Q.E.D.



My questions:



Could you please post a simpler solution?



I just don't understand why $frac 15+frac{ln 401}4<frac 74$. Please explain why is that.



Thank you in advance!






calculus inequality







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  • Just do it on you calculator; ($frac{1}{5} + frac{ln(401)}{4}) approx 1.7$ and $frac{7}{4} = 1.75$
    – Fiticous
    Dec 3 '18 at 6:45












  • $401<e^6$ then $ln(401)<6 implies frac{1}{5}+frac{ln(401)}{4}<frac{17}{10}=1.7 <frac{7}{4}$
    – Fareed AF
    Dec 3 '18 at 7:05


















2














Prove that
$ dfrac{1}{5}+ dfrac{1}{9}+ dfrac{1}{13}+ldots+ dfrac{ 1}{2005}< dfrac{ 7}{4}$.



Solution
$sum_{k=1}^{501}frac 1{4k+1}$ $=frac 15+sum_{k=2}^{501}frac 1{4k+1}$ $<frac 15+int_1^{501}frac{dx}{4x+1}$ $=frac 15+left[frac{ln(4x+1)}4right]_1^{501}$



So $sum_{k=1}^{501}frac 1{4k+1}$ $<frac 15+frac{ln 2005-ln 5}4$ $=frac 15+frac{ln 401}4$ $<frac 74$
Q.E.D.



My questions:



Could you please post a simpler solution?



I just don't understand why $frac 15+frac{ln 401}4<frac 74$. Please explain why is that.



Thank you in advance!






calculus inequality







share|cite|improve this question






















  • Just do it on you calculator; ($frac{1}{5} + frac{ln(401)}{4}) approx 1.7$ and $frac{7}{4} = 1.75$
    – Fiticous
    Dec 3 '18 at 6:45












  • $401<e^6$ then $ln(401)<6 implies frac{1}{5}+frac{ln(401)}{4}<frac{17}{10}=1.7 <frac{7}{4}$
    – Fareed AF
    Dec 3 '18 at 7:05
















2












2








2


2





Prove that
$ dfrac{1}{5}+ dfrac{1}{9}+ dfrac{1}{13}+ldots+ dfrac{ 1}{2005}< dfrac{ 7}{4}$.



Solution
$sum_{k=1}^{501}frac 1{4k+1}$ $=frac 15+sum_{k=2}^{501}frac 1{4k+1}$ $<frac 15+int_1^{501}frac{dx}{4x+1}$ $=frac 15+left[frac{ln(4x+1)}4right]_1^{501}$



So $sum_{k=1}^{501}frac 1{4k+1}$ $<frac 15+frac{ln 2005-ln 5}4$ $=frac 15+frac{ln 401}4$ $<frac 74$
Q.E.D.



My questions:



Could you please post a simpler solution?



I just don't understand why $frac 15+frac{ln 401}4<frac 74$. Please explain why is that.



Thank you in advance!






calculus inequality







share|cite|improve this question













Prove that
$ dfrac{1}{5}+ dfrac{1}{9}+ dfrac{1}{13}+ldots+ dfrac{ 1}{2005}< dfrac{ 7}{4}$.



Solution
$sum_{k=1}^{501}frac 1{4k+1}$ $=frac 15+sum_{k=2}^{501}frac 1{4k+1}$ $<frac 15+int_1^{501}frac{dx}{4x+1}$ $=frac 15+left[frac{ln(4x+1)}4right]_1^{501}$



So $sum_{k=1}^{501}frac 1{4k+1}$ $<frac 15+frac{ln 2005-ln 5}4$ $=frac 15+frac{ln 401}4$ $<frac 74$
Q.E.D.



My questions:



Could you please post a simpler solution?



I just don't understand why $frac 15+frac{ln 401}4<frac 74$. Please explain why is that.



Thank you in advance!






calculus inequality




calculus inequality






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share|cite|improve this question











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asked Dec 3 '18 at 6:38









nam

973




973












  • Just do it on you calculator; ($frac{1}{5} + frac{ln(401)}{4}) approx 1.7$ and $frac{7}{4} = 1.75$
    – Fiticous
    Dec 3 '18 at 6:45












  • $401<e^6$ then $ln(401)<6 implies frac{1}{5}+frac{ln(401)}{4}<frac{17}{10}=1.7 <frac{7}{4}$
    – Fareed AF
    Dec 3 '18 at 7:05




















  • Just do it on you calculator; ($frac{1}{5} + frac{ln(401)}{4}) approx 1.7$ and $frac{7}{4} = 1.75$
    – Fiticous
    Dec 3 '18 at 6:45












  • $401<e^6$ then $ln(401)<6 implies frac{1}{5}+frac{ln(401)}{4}<frac{17}{10}=1.7 <frac{7}{4}$
    – Fareed AF
    Dec 3 '18 at 7:05


















Just do it on you calculator; ($frac{1}{5} + frac{ln(401)}{4}) approx 1.7$ and $frac{7}{4} = 1.75$
– Fiticous
Dec 3 '18 at 6:45






Just do it on you calculator; ($frac{1}{5} + frac{ln(401)}{4}) approx 1.7$ and $frac{7}{4} = 1.75$
– Fiticous
Dec 3 '18 at 6:45














$401<e^6$ then $ln(401)<6 implies frac{1}{5}+frac{ln(401)}{4}<frac{17}{10}=1.7 <frac{7}{4}$
– Fareed AF
Dec 3 '18 at 7:05






$401<e^6$ then $ln(401)<6 implies frac{1}{5}+frac{ln(401)}{4}<frac{17}{10}=1.7 <frac{7}{4}$
– Fareed AF
Dec 3 '18 at 7:05












2 Answers
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3














This is not necessarily simpler because you have to know the upper bound on harmonic numbers $$H_n=sum_{k=1}^nfrac{1}{k}leq ln{n}+gamma+frac{1}{2n}$$ where $gammaapprox 0.577ldots$, but then you can write $$sum_{k=1}^{500}frac{1}{4k+1}<sum_{k=1}^{500}frac{1}{4k}=frac{1}{4}H_{500}leqfrac{1}{4}(ln{500}+gamma+frac{1}{1000})approx 1.70ldots<1.75$$






share|cite|improve this answer





























    3














    It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=frac{1}{4x+1}$ is convex over $mathbb{R}^+$,



    $$ int_{1/2}^{501+1/2}f(x),dx geq f(1)+f(2)+ldots+f(501) $$
    hence
    $$ sum_{k=1}^{501}frac{1}{4k+1}leq frac{log(669)}{4} $$
    and it is enough to show that $e^7>669.$ On the other hand $e>2+frac{2}{3}$ and
    $$left(2+frac{2}{3}right)^7=2^7cdotfrac{4}{3}cdotleft(2-frac{2}{9}right)^3=frac{2^{12}}{3}left(1-frac{1}{9}right)^3>frac{2^{13}}{3^2}>frac{8000}{9}>888.$$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






      active

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      active

      oldest

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      active

      oldest

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      3














      This is not necessarily simpler because you have to know the upper bound on harmonic numbers $$H_n=sum_{k=1}^nfrac{1}{k}leq ln{n}+gamma+frac{1}{2n}$$ where $gammaapprox 0.577ldots$, but then you can write $$sum_{k=1}^{500}frac{1}{4k+1}<sum_{k=1}^{500}frac{1}{4k}=frac{1}{4}H_{500}leqfrac{1}{4}(ln{500}+gamma+frac{1}{1000})approx 1.70ldots<1.75$$






      share|cite|improve this answer


























        3














        This is not necessarily simpler because you have to know the upper bound on harmonic numbers $$H_n=sum_{k=1}^nfrac{1}{k}leq ln{n}+gamma+frac{1}{2n}$$ where $gammaapprox 0.577ldots$, but then you can write $$sum_{k=1}^{500}frac{1}{4k+1}<sum_{k=1}^{500}frac{1}{4k}=frac{1}{4}H_{500}leqfrac{1}{4}(ln{500}+gamma+frac{1}{1000})approx 1.70ldots<1.75$$






        share|cite|improve this answer
























          3












          3








          3






          This is not necessarily simpler because you have to know the upper bound on harmonic numbers $$H_n=sum_{k=1}^nfrac{1}{k}leq ln{n}+gamma+frac{1}{2n}$$ where $gammaapprox 0.577ldots$, but then you can write $$sum_{k=1}^{500}frac{1}{4k+1}<sum_{k=1}^{500}frac{1}{4k}=frac{1}{4}H_{500}leqfrac{1}{4}(ln{500}+gamma+frac{1}{1000})approx 1.70ldots<1.75$$






          share|cite|improve this answer












          This is not necessarily simpler because you have to know the upper bound on harmonic numbers $$H_n=sum_{k=1}^nfrac{1}{k}leq ln{n}+gamma+frac{1}{2n}$$ where $gammaapprox 0.577ldots$, but then you can write $$sum_{k=1}^{500}frac{1}{4k+1}<sum_{k=1}^{500}frac{1}{4k}=frac{1}{4}H_{500}leqfrac{1}{4}(ln{500}+gamma+frac{1}{1000})approx 1.70ldots<1.75$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 7:16









          Michal Adamaszek

          2,08148




          2,08148























              3














              It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=frac{1}{4x+1}$ is convex over $mathbb{R}^+$,



              $$ int_{1/2}^{501+1/2}f(x),dx geq f(1)+f(2)+ldots+f(501) $$
              hence
              $$ sum_{k=1}^{501}frac{1}{4k+1}leq frac{log(669)}{4} $$
              and it is enough to show that $e^7>669.$ On the other hand $e>2+frac{2}{3}$ and
              $$left(2+frac{2}{3}right)^7=2^7cdotfrac{4}{3}cdotleft(2-frac{2}{9}right)^3=frac{2^{12}}{3}left(1-frac{1}{9}right)^3>frac{2^{13}}{3^2}>frac{8000}{9}>888.$$






              share|cite|improve this answer


























                3














                It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=frac{1}{4x+1}$ is convex over $mathbb{R}^+$,



                $$ int_{1/2}^{501+1/2}f(x),dx geq f(1)+f(2)+ldots+f(501) $$
                hence
                $$ sum_{k=1}^{501}frac{1}{4k+1}leq frac{log(669)}{4} $$
                and it is enough to show that $e^7>669.$ On the other hand $e>2+frac{2}{3}$ and
                $$left(2+frac{2}{3}right)^7=2^7cdotfrac{4}{3}cdotleft(2-frac{2}{9}right)^3=frac{2^{12}}{3}left(1-frac{1}{9}right)^3>frac{2^{13}}{3^2}>frac{8000}{9}>888.$$






                share|cite|improve this answer
























                  3












                  3








                  3






                  It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=frac{1}{4x+1}$ is convex over $mathbb{R}^+$,



                  $$ int_{1/2}^{501+1/2}f(x),dx geq f(1)+f(2)+ldots+f(501) $$
                  hence
                  $$ sum_{k=1}^{501}frac{1}{4k+1}leq frac{log(669)}{4} $$
                  and it is enough to show that $e^7>669.$ On the other hand $e>2+frac{2}{3}$ and
                  $$left(2+frac{2}{3}right)^7=2^7cdotfrac{4}{3}cdotleft(2-frac{2}{9}right)^3=frac{2^{12}}{3}left(1-frac{1}{9}right)^3>frac{2^{13}}{3^2}>frac{8000}{9}>888.$$






                  share|cite|improve this answer












                  It is enough to invoke the Hermite-Hadamard inequality. Since $f(x)=frac{1}{4x+1}$ is convex over $mathbb{R}^+$,



                  $$ int_{1/2}^{501+1/2}f(x),dx geq f(1)+f(2)+ldots+f(501) $$
                  hence
                  $$ sum_{k=1}^{501}frac{1}{4k+1}leq frac{log(669)}{4} $$
                  and it is enough to show that $e^7>669.$ On the other hand $e>2+frac{2}{3}$ and
                  $$left(2+frac{2}{3}right)^7=2^7cdotfrac{4}{3}cdotleft(2-frac{2}{9}right)^3=frac{2^{12}}{3}left(1-frac{1}{9}right)^3>frac{2^{13}}{3^2}>frac{8000}{9}>888.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 12:43









                  Jack D'Aurizio

                  286k33279656




                  286k33279656






























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