無限級数その１

［証明］
[11]　0$<$$r$$<$1のとき，
　$$\sum _{n=1}^N\frac{r^n}{n}=\sum _{n=1}^N{r}^n{\int }_0^1{x}^{n-1}dx=r{\int }_0^1\frac{1-(rx{)}^N}{1-rx}dx$$
ここで，$N$$\to$0のとき，
$$\left|{\int }_0^1\frac{(rx{)}^N}{1-rx}dx\right|\leqq \left|{\int }_0^1\frac{x^N}{1-r}dx\right|\leqq \frac{1}{N+1}\frac{1}{1-r}\to 0$$
したがって，
$$\sum _{n=1}^N\frac{r^n}{n}=\to r{\int }_0^1\frac{1}{1-rx}dx$$
$$r{\int }_0^1\frac{1}{1-rx}dx={\int }_0^1\frac{1}{\frac{1}{r}-x}dx=-\log (1-r)．\square \square$$
[12]-1$<$$r$$<$0のとき，0$<$$r^2<$1から，
$$\sum _{n=1}^{2N}\frac{r^n}{n}+\sum _{n=1}^{2N}\frac{(-r{)}^n}{n}=2\sum _{n=1}^N\frac{r^{2n}}{2n}=\sum _{n=1}^N\frac{r^{2n}}{n}$$
$$\sum _{n=1}^{2N}\frac{r^n}{n}=-\sum _{n=1}^{2N}\frac{(-r{)}^n}{n}+\sum _{n=1}^N\frac{r^{2n}}{n}$$
$$\sum _{n=1}^{2N}\frac{r^n}{n}=\to \log (1+r)-\log (1-r^2)=-\log (1-r)．\square \square$$
[13]$r=-1$のとき，
　$$\sum _{n=1}^N\frac{(-1{)}^n}{n}=\sum _{n=1}^N(-1{)}^n{\int }_0^1{x}^{n-1}dx=-{\int }_0^1\frac{1-(-x{)}^N}{1+x}dx$$
ここで，$N$$\to$0のとき，
$$\left|{\int }_0^1\frac{(-x{)}^N}{1+x}dx\right|\leqq \left|{\int }_0^1{x}^{N}dx\right|\leqq \frac{1}{N+1}\to 0$$
したがって，
$$\sum _{n=1}^N\frac{(-1{)}^n}{n}=\to -{\int }_0^1\frac{1}{1+x}dx=-\log 2．\square \square$$
以上から，定理01は成り立つ．□□

[21]　0$<$$r$$<$1のとき，　$$\sum _{n=0}^{N-1}\frac{r^n}{1+Dn}=\sum _{n=0}^{N-1}{r}^n{\int }_0^1{x}^{Dn}dx={\int }_0^1\frac{1-(r{x}^D{)}^N}{1-r{x}^D}dx$$
ここで，$N$$\to$0のとき，
$$\left|{\int }_0^1\frac{(r{x}^D{)}^N}{1-r{x}^D}dx\right|\leqq \left|{\int }_0^1\frac{x^{DN}}{1-r}dx\right|\leqq \frac{1}{DN+1}\frac{1}{1-r}\to 0$$
したがって，
$$\sum _{n=0}^{N-1}\frac{r^n}{1+Dn}=\to {\int }_0^1\frac{1}{1-r{x}^D}dx．\square \square$$
[22]-1$<$$r$$<$0のとき，0$<$$r^2<$1から，
$$\sum _{n=0}^{2N-1}\frac{r^n}{1+Dn}=\sum _{n=0}^{N-1}\frac{r^{2n}}{1+(2n)D}+\sum _{n=0}^{N-1}\frac{r^{2n+1}}{1+(2n+1)D}$$
$$\sum _{n=0}^{N-1}\frac{r^{2n}}{1+(2n)D}\to {\int }_0^1\frac{1}{1-r^2{x}^{2D}}dx．\square \square$$
$$\sum _{n=0}^{N-1}\frac{r^{2n+1}}{1+(2n+1)D}=\frac{r}{1+D}\sum _{n=0}^{N-1}\frac{r^{2n}}{1+n(\frac{2D}{D+1})}$$
$$\to \frac{r}{1+D}{\int }_0^1\frac{1}{1-r^2{x}^{\frac{2D}{1+D}}}dx．$$
$$=\frac{r}{1+D}(1+D){\int }_0^1\frac{y^D}{1-r^2{y}^{2D}}dy={\int }_0^1\frac{r{x}^D}{1-r^2{x}^{2D}}dx．$$
$$\sum _{n=0}^{2N-1}\frac{r^n}{1+Dn}\to {\int }_0^1\frac{1}{1-r^2{x}^{2D}}dx+{\int }_0^1\frac{r{x}^D}{1-r^2{x}^{2D}}dx．$$
$$={\int }_0^1\frac{1}{1-r{x}^D}dx．$$

[23]$r=-1$のとき，
$$\sum _{n=0}^{2N-1}\frac{(-1{)}^n}{1+Dn}=\sum _{n=0}^{N-1}\frac{1}{1+(2n)D}-\sum _{n=0}^{N-1}\frac{1}{1+(2n+1)D}$$
$$={\int }_0^1\sum _{n=0}^{N-1}(x^{2nD}-x^{(2n+1)D})dx$$
$$={\int }_0^1(1-x^D)\frac{1-(x^{2D}{)}^N}{1-x^{2D}}dx={\int }_0^1\frac{1}{1+x^D}dx-{\int }_0^1\frac{x^{2DN}}{1+x^D}dx$$

ここで，$N$$\to$0のとき，
$$\left|{\int }_0^1\frac{x^{2DN}}{1+x^D}dx\right|\leqq \left|{\int }_0^1{x}^{2DN}dx\right|\leqq \frac{1}{2DN+1}\to 0$$
したがって，
$$\sum _{n=0}^{N-1}\frac{(-1{)}^n}{1+Dn}={\int }_0^1\frac{1}{1+x^D}dx．\square \square$$
以上から，定理02は成り立つ．□□